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Vaishnavi Gadhe
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in Quantitative Aptitude
Jul 3, 2022

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1 vote

**Answer - 3. 8**

Let a be the first term and r be the common ration of given GP –

GP : a, ar, ar^2, ar^3, ar^4, …

Sum of first and third term is 40 ie a + ar^2 = 40

1. a(1 + r^2) = 40.

Sum of second and fourth term is 80 ie ar + ar^3 = 80

2. ar(1 + r^2) = 80.

From 1 and 2 - r (40) = 80, therefore r = 2.

From 1 - a(1 + 2^2) = 40, therefore a = 8.

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Option: C

let, first term of G.P be “a” and common difference be “r”

i.e., 1^{st} term = a, 2^{nd} term = ar, 3^{rd} term = ar^2, 4^{th} term = ar^3.

given, sum of 1^{st} and 2^{nd} term = 40

a + ar^2 = 40 ----→ let it be equation 1.

and, sum of 2^{nd} and 4^{th} term = 80

ar + ar^3 = 80 -----→ let it be equation 2.

r(ar+ar^2) = 80 -------→ substitute equation 1 in equation 2

r (40) = 80

r = 80/40 = 2

r = 2

i.e., substitute r value in equation 1 or equation 2 to get the a value.

substituting in equation 1, gives:

a+a(2)^2 = 40

a+4a=40

5a=40

a=40/5

a=8.

Therefore the first term of Geometric preogression is 8.