Question

If for G.P. r=2, S10 = 1023 then a =
a)-1
b) 1
c) 2
d) -2​

Answer 1

Given:

r = 2 and $S_{10}$ = 1023

We have to find, the value of a.

Solution:

We know that:

The sum of nth term of a GP

$S_{n} =\dfrac{a(r^{n} -1)}{r-1}$ , since r > 1

$\therefore$ $S_{10} =\dfrac{a(2^{10} -1)}{2-1}=1023$

⇒ $\dfrac{a(1024 -1)}{1}=1023$

⇒ a(1023) = 1023

⇒ a = 1

Thus, the required "option b) 1" is correct.

Answer 2

Given :- If for G.P. r=2, S10 = 1023 then a =

a)-1

b) 1

c) 2

d) -2

Solution :-

we know that,

• sum of n terms of GP = a[r^n - 1] / (r - 1) where
• a = first term = Let a
• r = common ratio = 2
• n = total terms = 10 .

putting all values we get,

→ a[r^n - 1] / (r - 1) = 1023

→ a(2^10 - 1) / (2 - 1) = 1023

→ a(2^10 - 1) = 1023

→ a(1024 - 1) = 1023

→ a * 1023 = 1023

→ a = 1 (b) (Ans.)

Learn more :-

Verify whether the given sequence are in H.P 1/3,1/6,1/12,1/24....

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