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Determine the first term of geometric progression is the sum of first term and third term is 40 and sum of second term and fourth term  is 80

1)12

2)16

3)8

4)4

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.

Option: C

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

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

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

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

and, sum of 2nd and 4th 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.