The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the GP and the first term of the G.P. is equal to common difference of the A.P., then the sum

Question

The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the GP and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is: [2026]

Smart Achievers · Accepted Answer

(1)

Sum of first 10 terms of A.P. $S_{10} = 160$

$\Rightarrow 5 [2 a + 9 d] = 160 \Rightarrow 2 a + 9 d = 32$

First term of G.P. = Common difference of A.P. = d

and ratio of G.P. = First term of A.P.

So, G.P. is d, da, $d a^{2}$ ...

Now, d + da = 8

$\Rightarrow d + d \frac{(32 - 9 d)}{2} = 9$

$\Rightarrow 2 d + 32 d - 9 d^{2} = 16$

$\Rightarrow 9 d^{2} = 34 d + 16 = 0$

Required sum $= d_{1} + d_{2} = \frac{34}{9}$

Solution

1 $\frac{34}{9}$

2 $\frac{34}{13}$

3 $\frac{32}{9}$

4 $\frac{32}{13}$

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Solution

1 349

2 3413

3 329

4 3213

Ans. (1) Sum of first 10 terms of A.P. S10=160 ⇒ 5[2a+9d]=160 ⇒ 2a+9d=32 First term of G.P. = Common difference of A.P. = d and ratio of G.P. = First term of A.P. So, G.P. is d, da, da2... Now, d + da = 8 ⇒ d+d(32–9d)2=9 ⇒ 2d+32d–9d2=16 ⇒ 9d2=34d+16=0 Required sum =d1+d2=349

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1 $\frac{34}{9}$

2 $\frac{34}{13}$

3 $\frac{32}{9}$

4 $\frac{32}{13}$