ARITHMETIC PROGRESSIONS-Exercise 5.1

Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

1. The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
2. The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
3. The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
4. The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8 % per annum.

Solution: (i)

The taxi fare structure begins with ₹15 for the first kilometer. For each additional kilometer traveled, an extra ₹8 is charged. Thus, the total fare after completing exactly n kilometers forms a sequence where the first term is ₹15, and each subsequent term increases by ₹8.

The fares accumulate as follows: after 1 km, the fare is ₹15; after 2 km, it becomes ₹15 + ₹8 = ₹23; after 3 km, ₹15 + ₹8 + ₹8 = ₹31; and so on. This pattern continues indefinitely for each additional kilometer.

The sequence of fares can be expressed mathematically as:

This clearly forms an Arithmetic Progression (AP) where the first term and common difference are given by:

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Solution: (ii)

Air Amount in Cylinder - Not an AP

The vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder each time, leaving \(\frac{3}{4}\) of the previous amount after every operation. If the initial volume of air is \(V\), then after the first pump, \(\frac{3}{4}V\) remains; after the second, \(\frac{3}{4} \times \frac{3}{4}V = \left(\frac{3}{4}\right)^2 V\); after the third, \(\left(\frac{3}{4}\right)^3 V\), and so on.

This creates a sequence: \(V, \ \frac{3}{4}V, \ \left(\frac{3}{4}\right)^2 V, \ \left(\frac{3}{4}\right)^3 V, \ \ldots\)

This sequence forms a Geometric Progression with first term \(a = V\) and common ratio \(r = \frac{3}{4}\), not an Arithmetic Progression. In an AP, the difference between consecutive terms remains constant, but here the amount removed decreases each time (\(\frac{1}{4}V, \ \frac{3}{16}V, \ \frac{9}{64}V, \ \ldots\)), showing a constant ratio instead of a constant difference.

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Solution: (iii)

Digging Cost as Arithmetic Progression

The cost of digging a well starts at ₹150 for the first meter and increases by ₹50 for each subsequent meter. This means the cost per meter forms a sequence where the first meter costs ₹150, the second meter costs ₹150 + ₹50 = ₹200, the third meter ₹250, and continues rising by ₹50 each time.

The sequence of costs per meter is expressed as:

This pattern clearly represents an Arithmetic Progression where each term increases by a fixed amount. The first term represents the initial digging cost while the common difference captures the rising expense for deeper meters.

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Solution: (iv)

Compound Interest - Geometric Progression

The initial deposit of ₹10,000 grows at 8% compound interest per annum, meaning each year the amount increases by a factor of 1.08. After the first year, the amount becomes ₹10,000 × 1.08 = ₹10,800; after the second year, ₹10,800 × 1.08 = ₹11,664; after the third year, ₹11,664 × 1.08 = ₹12,597.12, and this multiplication continues annually.

This forms a sequence where each term is obtained by multiplying the previous term by 1.08:

The pattern represents a Geometric Progression with the initial principal as the first term and 1.08 as the constant multiplication factor each year. Unlike arithmetic progression where addition is constant, here the growth is multiplicative, characteristic of compound interest calculations.

Q2. Write first four terms of the AP, when the first term a and the common difference d are given as follows:

1. \(a = 10, d = 10\)
2. \( a = –2, d = 0\)
3. \(a = 4, d = – 3\)
4. \( a = – 1, d = \frac{1}{2}\)
5. \( a = – 1.25, d = – 0.25\)

Solution: (i)

Given:
\(a=10,\; d=10\) $$\begin{aligned} a_{1}&=10\\\\ a_{2}&=a_{1}+d\\&=10+10\\&=20\\\\ a_{3}&=a_{1}+2d\\&=10+2\times 10\\&=30\\\\ a_{4}&=a_{1}+3d\\&=10\times 3\times 10\\&=40\end{aligned}$$ First four terms of AP
10, 20, 30, 40

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Solution: (ii)

Given:
\(a=-2,\; d=0\) $$\begin{aligned} a_{1}&=-2\\\\ a_{2}&=a_{1}+d\\&=-2+0\\&=-2\\\\ a_{3}&=a_{1}+2d\\&=-2+2\cdot 0\\&=-2\\\\ a_{4}&=a_{1}+3d\\&=-2+3\cdot 0\\&=-2\end{aligned}$$ First four terms of AP
-2, -2, -2, -2

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Solution: (iii)

Given:
\(a=4,\; d=-3\) $$\begin{aligned} a_{1}&=4\\\\ a_{2}&=a_{1}+d\\ &=4-3\\ &=1\\\\ a_{3}&=a_{1}+2d\\ &=4+\left( 2\times \left( -3\right) \right) \\ &=4-6\\ &=-2\\\\ a_{4}&=a_{1}+3d\\ &=4+\left( 3\times \left( -3\right) \right) \\ &=4-9\\ \\ &=-5\end{aligned}$$ First four terms of AP
4, 1, -2, -5

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Solution: (iv)

Given:
\(a=-1,\; d=\dfrac{1}{2}\) $$\begin{aligned} a_{1}&=-1\\\\ a_{2}&=a_{1}+d\\ &=-1+\dfrac{1}{2}\\ &=-\dfrac{1}{2}\\\\ a_{3}&=a_{1}+2d\\ &=-1+\left( 2\cdot \dfrac{1}{2}\right) \\ &=0\\\\ a_{4}&=a_{1}+3d\\ &=-1+\left( 3\cdot \dfrac{1}{2}\right) \\ &=\dfrac{1}{2}\end{aligned}$$

First four terms of AP
-1, \(-\frac{1}{2}\), 0, \(\frac{1}{2}\)

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Solution: (v)

Given:
\(a=-1.25,\; d=-0.25\) $$\begin{aligned} a_{1}&=-1.25\\\\ a_{2}&=a_{1}+d\\ &=-1.25-0.25\\ &=1.50\\\\ a_{3}&=a_{1}+2d\\ &=-1.25-\left( 2\times (0.25)\right) \\ &=-1.25-0.50\\ &=-1.75\\\\ a_{4}&=a_{1}+2d\\ &=-1.25-\left( 3\times (0.25)\right) \\ &=-1.25-0.75\\ &=-2.0\end{aligned}$$ First four terms of AP
-1.25, 1.50, 1.75, 2.0

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Q3. For the following APs, write the first term and the common difference:

1. \(3,\ 1, – 1,\ – 3,\ . . .\)
2. \(– 5,\ – 1,\ 3,\ 7,\ . . .\)
3. \(\frac{1}{3},\ \frac{5}{3},\ \frac{9}{3},\ \frac{13}{3},\ \)
4. \(0.6,\ 1.7,\ 2.8,\ 3.9,\ . . .\)

Solution: (i)

Given:
\(3,\ 1,\ -1,\ -3\ ....\) $$ \begin{aligned} a_1&=3,\\a_2&=1\\\\ d&=a_{2}-a_{1}\\ &=1-3\\ &=-2 \end{aligned}$$

First term \(a_1=3\)
Common Difference \(d=-2\)

Solution: (ii)

Given:
\(-5,\ -1,\ 3,\ 7,\ \ldots\) $$ \begin{aligned}a_1&=-5\\ a_2&=-1\\\\ d&=a_{2}-a_{1}\\ &=-1-\left( -5\right)\\ &=-1+5\\ &=4 \end{aligned} $$

First term \(a_1=-5\)
Common Difference \(d=4\)

Solution: (iii)

Given:
\(\dfrac{1}{3},\ \dfrac{5}{3},\ \dfrac{9}{3},\ \dfrac{13}{3},\ \ldots\) $$ \begin{aligned} a_1&=\dfrac{1}{3}\\ a_2&=\dfrac{5}{3}\\\\ d&=a_{2}-a_{1}\\ &=\dfrac{5}{3}-\dfrac{1}{3}\\ &=\dfrac{4}{3} \end{aligned} $$

First term \(a_1=\frac{1}{3}\)
Common Difference \(d=\dfrac{4}{3}\)

Solution: (iv)

Given:
\(0.6,\ 1.7,\ 2.8,\ 3.9,\ \ldots\) $$ \begin{aligned} a_1&=0.6\\ a_2&=1.7\\\\ d&=a_{2}-a_{1}\\ &=1.7-0.6\\ &=1.1 \end{aligned} $$

First term \(a_1=0.6\)
Common Difference \(d=1.1\)

Q4. Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

1. \( 2,\ 4,\ 8,\ 16,\ \ldots \)
2. \(2,\ \frac{5}{2},\ 3,\ \frac{7}{2},\ \ldots\)
3. \( – 1.2,\ – 3.2,\ – 5.2,\ – 7.2,\ \ldots \)
4. \(– 10,\ – 6,\ – 2,\ 2,\ \ldots\)
5. \(3,\ 3 + \sqrt{2} ,\ 3 + 2 \sqrt{2} ,\ 3 + 3 \sqrt{2},\ \ldots \)
6. \(0.2,\ 0.22,\ 0.222,\ 0.2222,\ \ldots\)
7. \(0,\ – 4,\ – 8,\ –12,\ \ldots \)
8. \(-\frac{1}{2},\ -\frac{1}{2},\ -\frac{1}{2},\ -\frac{1}{2},\ \ldots\)
9. \(1,\ 3,\ 9,\ 27,\ \ldots \)
10. \( a,\ 2a,\ 3a,\ 4a,\ \ldots\)
11. \(a,\ a^2,\ a^3,\ a^4,\ \ldots\)
12. \(\sqrt{2},\ \sqrt{8},\ \sqrt{18},\ \sqrt{32},\ \ldots \)
13. \(\sqrt{3},\ \sqrt{6},\ \sqrt{9},\ \sqrt{12},\ \ldots \)
14. \(1^2,\ 5^2,\ 5^2,\ 7^2,\ \ldots \)
15. \(1^2,\ 5^2,\ 7^2,\ 73,\ \ldots \)

Solution: (i)

Given:
\(2,\ 4,\ 8,\ 16,\ \ldots\) $$ \begin{aligned}d&=a_{2}-a_{1}\\ &=4-2\\ &=2\\\\ d&=a_{3}-a_{2}\\ &=8-4\\ &=4\\\\ d=a_{2}-a_{1}&\ne a_{3}-a_{2} \end{aligned}$$

Common difference \(d\) is not equal therefore, given series is not in AP

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Solution: (ii)

Given:
\(2,\ \dfrac{5}{2},\ 3,\ \dfrac{7}{2},\ \ldots\) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=\dfrac{5}{2}-2\\ &=\dfrac{5-4}{2}\\ &=\dfrac{1}{2}\\\\ d&=a_{3}-a_{2}\\ &=3-\dfrac{5}{2}\\ &=\dfrac{6-5}{2}\\ &=\dfrac{1}{2}\\\\ d&=a_{4}-a_{3}\\ &=\dfrac{7}{2}-3\\ &=\dfrac{7-6}{2}\\ &=\dfrac{1}{2}\\ \end{aligned}$$

GIven Series is in an AP with Common Difference \(d=\dfrac{1}{2}\)
First term \(a_1=2\)
therefore,

$$\begin{aligned} a_{5}&=a_{1}+4d\\ &=2+4\cdot \dfrac{1}{2}\\ &=2+2\\ &=4\\\\ a_{6}&=a_{1}+5d\\ &=2+5\cdot \dfrac{1}{2}\\ &=\dfrac{4+5}{2}\\ &=\dfrac{9}{2}\\\\ a_{7}&=a_{1}+6d\\ &=2+6\cdot \dfrac{1}{2}\\ &=5\end{aligned}$$ Series with additional terms \(2,\ \dfrac{5}{2},\ 3,\ \dfrac{7}{2},\ 4,\ \dfrac{9}{2},\ 5,\ \ldots\)

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Solution: (iii)

Given:
\(-1.2,\ -3.2,\ -5.2,\ -7.2,\ \ldots\) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=-3.2-\left( -1.2\right) \\ &=-3.2+1.2\\ &=-2\\\\ d&=a_{3}-a_{1}\\ &=-5.2-\left( -3.2\right) \\ &=-5.2+3.2\\ &=-2\\\\ d&=a_{4}-a_{3}\\ &=-7.2-\left( -5.2\right) \\ &=-7.2+5.2\\ &=-2\end{aligned}$$

Series in an AP
First term \(a=-1.2\)
Common difference \(d=-2\)

$$\begin{aligned} a_{5}&=a_{1}+4d\\ &=-1.2+\left( 4\cdot \left( -2\right) \right) \\ &=-1.2-8\\ &=-9.2\\\\ a_{6}&=a_{1}+5d\\ &=-1.2+\left( 5\cdot \left( -2\right) \right) \\ &=-1.2-10\\ &=-11.2\\\\ a_{7}&=a_{1}+6d\\ &=-1.2+\left( 6\cdot \left( -2\right) \right) \\ &=-1.2-12\\ &=-13.2\end{aligned}$$ Series with additional 3 terms \(-1.2,\ -3.2,\ -5.2,\ -7.2,\ -9.2,\ -11.2,\ -13.2,\ \ldots\)

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Solution: (iv)

Given:
-10, -6, -2, 2, .... $$\begin{aligned} d&=a_{2}-a_{1}\\ &=-6-\left( -10\right) \\ &=-6+10\\ &=4\\\\ d&=a_{3}-a_{2}\\ &=-2-\left( -6\right) \\ &=-2+6\\ &=4\\\\ d&=a_{4}-a_{3}\\ &=2-\left( -2\right) \\ &=4\end{aligned}$$

Series is in AP with Common Difference \(d=4\)
First term \(a_1=-10\)

$$\begin{aligned} a_{5}&=a_{1}+4a\\ &=-10+\left( 4\times 4\right) \\ &=-10+16\\ &=6\\\\ a_{6}&=a_{1}+5d\\ &=-10+\left( 5\times 4\right) \\ &=-10+20\\ &=10\\\\ a_{7}&=a_{1}+6d\\ &=-10+\left( 6\times 4\right) \\ &=-10+24\\ &14\end{aligned}$$ Series with additional 3 terms -10, -6, -2, 2, 6, 10, 14, ....

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Solution: (v)

Given:
\(3,\ 3+\sqrt{2},\ 3+2\sqrt{2},\ 3+3\sqrt{2},\ ...\) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=3+\sqrt{2}-3\\ &=\sqrt{2}\\\\ d&=a_{3}-a_{2}\\ &=3+2\sqrt{2}-3-\sqrt{2}\\ &=\sqrt{2}\\\\ d&=a_{4}-a_{3}\\ &=3+3\sqrt{2}-3-2\sqrt{2}\\ &=\sqrt{2}\end{aligned}$$

Series in AP with Common Difference \(d=\sqrt{2}\) First term \(a_1=3\)

$$\begin{aligned} a_{5}&=a_{1}+4d\\ &=3+4\sqrt{2}\\\\ a_{6}&=a_{1}+5d\\ &=3+5\sqrt{2}\\\\ a_{7}&=a_{1}+6d\\ &=3+6\sqrt{2}\end{aligned}$$ Series with additional 3 terms \(3,\ 3+\sqrt{2},\ 3+2\sqrt{2},\ 3+3\sqrt{2},\ 3+4\sqrt{2},\ 3+5\sqrt{2},\ 3+6\sqrt{2},\ ...\)

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Solution: (vi)

Given:
\(0.2,\ 0.22,\ 0.222,\ 0.222,\ \ldots \) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=0.22-0.2\\ &=0.02\\\\ d&=a_{3}-a_{2}\\ &=0.222-0.22\\ &=0.002 \end{aligned}$$ \[d=a_{2}-a_{1}\neq a_{3}-a_{2}\] Hence series is not in AP.

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Solution: (vii)

Given:
\(0,\ -4,\ -8,\ -12,\ \ldots\) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=-4-0\\ &=-4\\\\ d&=a_{3}-a_{2}\\ &=-8-\left( -4\right) \\ &=-8+4\\ &=-4\\\\ d&=a_{4}-a_{3}\\ &=-12-\left( -8\right) \\ &=-12+8\\ &=-4\end{aligned}$$

Series is in AP with Common Difference \(d=-4\)
First term \(a_{1}=0\)

$$\begin{aligned} a_{5}&=a_{1}+4d\\ &=0+\left( 4\times(-4)\right) \\ &=-16\\\\ a_{6}&=a_{1}+5d\\ &=0-4\times 5\\ &=-20\\\\ a_{7}&=a_{1}+6d\\ &=0-6\times 4\\ &=-24\end{aligned}$$ Series with additional 3 terms \(0,\ -4,\ -8,\ -12,\ -16,\ -20,\ -24,\ \ldots\)

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Solution: (viii)

Given:
$$-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2}...$$

All the terms are same, therefore, Common Difference \(d=0\) First term \(a_1=-\dfrac{1}{2}\)

$$a=-\dfrac{1}{2}$$

Since \(d=0\), all the terms will be same as and it is in AP.
addition 3 terms of the series is as following

$$-\dfrac{1}{2},\ -\dfrac{1}{2},\ -\dfrac{1}{2}$$

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Solution: (ix)

Given:
\(1,\ 3,\ 9,\ 7,\ \ldots \) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=3-1\\ &=2\\\\ d&=a_{3}-a_{2}\\ &=9-3\\ &=6\\\\ d&=a_{4}-a_{3}\\ &=27-9=\\ &18\\ \end{aligned}$$ Since, \[d=a_{2}-a_{1}\ne a_{3}-a_{2} \ne a_{4}-a_{3}\] Series is not in AP

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Solution: (x)

Given:
\(a,\ 2a,\ 3a,\ 4a\ ...\) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=2a-a\\ &=a\\\\ d&=a_{3}-a_{2}\\ &=3a-2a\\ &=a\\\\ d&=a_{4}-a_{3}\\ &=4a-3a\\ &=a\end{aligned}$$

series is in AP with Common Difference \(d=a\),
First term \(a_{1}=a\)

$$\begin{aligned} a_{5}&=a_{1}+4d\\ &=a+4a\\ &=5a\\\\ a_{6}&=a_{1}+5d\\ &=a+5a\\ &=6a\\\\ a_{7}&=a_{1}+6d\\ &=a+6a\\ &=7a\end{aligned}$$ Series with additional 3 terms \(a,\ 2a,\ 3a,\ 4a,\ 5a,\ 6a,\ 7a,\ ...\)

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Solution: (xi)

Given:
\(a,\ a^{2},\ a^{3},\ a^{4},\ \ldots \) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=a^{2}-a\\ &=a\left( a-1\right) \\\\ d&=a_{3}-a_{2}\\ &=a^{3}-a^{2}\\ &=a^{2}\left( a-1\right) \\\\ d&=a_{4}-a_{3}\\ &=a^{4}-a^{3}\\ &=a^{3}\left( a-1\right) \\ \end{aligned}$$

Common Difference \(d=a_{2}-a_{1}\ne a_{3}-a_{2} \ne a_{4}-a_{3}\), Hence, given series is not in AP

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Solution: (xii)

Given:
\(\sqrt{2},\ \sqrt{8},\ \sqrt{18},\ \sqrt{32},\ \ldots\) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=\sqrt{8}-\sqrt{2}\\ &=2\sqrt{2}-\sqrt{2}\\ &=\sqrt{2}\\\\ d&=a_{3}-a_{2}\\ &=\sqrt{18}-\sqrt{8}\\ &=3\sqrt{2}-2\sqrt{2}\\ &=\sqrt{2}\\\\ d&=a_{4}-a_{3}\\ &=\sqrt{32}-\sqrt{18}\\ &=4\sqrt{2}-3\sqrt{2}\\ &=\sqrt{2}\\\\ d=\sqrt{2}=\sqrt{2}=\sqrt{2}\end{aligned}$$

series is in AP with Common Difference \(d=\sqrt{2}\),br< First term \(a_{1}=\sqrt{2}\)

$$\begin{aligned} a_{5}&=a_{1}+4d\\ &=\sqrt{2}+452\\ &=\sqrt{2}\left( 1+4\right) \\ &=5\sqrt{2}\\ &=\sqrt{50}\\\\ a_{6}&=a_{1}+5d\\ &=\sqrt{2}+552\\ &=\sqrt{2}\left( 1+5\right) \\ &=6\sqrt{2}\\ &=\sqrt{72}\\\\ a_{7}&=a_{1}+6d\\ &=\sqrt{2}+6\sqrt{2}\\ &=7\sqrt{2}\\ &=\sqrt{98}\end{aligned}$$ Series with additional 3 terms \(\sqrt{2},\ \sqrt{8},\ \sqrt{18},\ \sqrt{32},\ \sqrt{50},\ \sqrt{72},\ \sqrt{98},\ \ldots\)

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Solution: (xiii)

Given:
\(\sqrt{3},\ \sqrt{6},\ \sqrt{9},\ \sqrt{12},\ \ldots \) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=\sqrt{6}-\sqrt{3}\\\\ d&=a_{3}-a_{2}\\ &=\sqrt{9}-\sqrt{6}\\ &=3-\sqrt{6}\\\\ d&=a_{4}-a_{3}\\ &=\sqrt{12}-\sqrt{9}\\ &=2\sqrt{3}-3\\ \end{aligned}$$

Common Diffrenece \(d=a_{2}-a_{1}\ne a_{3}-a_{2} \ne a_{4}-a_{3}\), Hence series is not in AP

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Solution: (xiv)

Given:
\(1^{2},\ 3^{2},\ 5^{2},\ 7^{2}...\) $$ \begin{aligned} d&=a_{2}-a_{1}\\ &=3^{2}-1^{2}\\ &=9-1\\ &=8\\\\ d&=a_{3}-a_{2}\\ &=5^{2}-3^{2}\\ &=25-9\\ &=16\\\\ d&=a_{4}-a_{3}\\ &=7^{2}-5^{2}\\ &=49-25\\ &=24 \end{aligned} $$

Common Difference \(d=a_{2}-a_{1} \ne a_{3}-a_{2}\ne a_{4}-a_{3}\), Hence, series is not in AP

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Solution: (xv)

Given:
\(1^{2},\ 5^{2},\ 7^{2},\ 73,\ \ldots \) $$\begin{aligned} d&=a_{2}-a_{1}\\ &=5^{2}-1^{2}\\ &=25-1\\ &=24\\\\ d&=a_{3}-a_{2}\\ &=7^{2}-5^{2}\\ &=49-25\\ &=24\\\\ d&=a_{4}-a_{3}\\ &=73-7^{2}\\ &=73-49\\ &=24\\ d=24=24=24\end{aligned}$$

Hence series is in AP with Common Difference \(d=24\)
First term \(a_{1}=1\)

$$\begin{aligned} a_{5}&=a_{1}+4d\\ &=1+4\times 24\\ &=1+96\\ &=97\\\\ a_{6}&=a_{1}+5d\\ &=1+5\times 24\\ &=1+120\\ &=121\\\\ a_{7}&=a_{1}+6d\\ &=1+6\times 24\\ &=1+144\\ &=145\end{aligned}$$ Series with additional 3 terms \(1^{2},\ 5^{2},\ 7^{2},\ 73,\ 97,\ 11^2,\ 145,\ \ldots \)