Question

help the circled one pity clear

Answer 1

Hi,

Sn = n/2[2a+(n-1)d]
Sn₁ = n/2[2a'+(n-1)d']

Given Ratio of the Sum of n Terms of two APs = (2n+1) : (2n+5)

Sn/Sn' = (2n+1) / (2n+5)

n/2[2a+(n-1)d] / n/2[2a'+(n-1)d'] = (2n+1) / (2n+5)

[2a+(n-1)d] / [2a'+(n-1)d'] = (2n+1) / (2n+5)      ( ÷2 on LHS)

[a+(n-1/2)d] / [a'+(n-1/2)d'] = (2n+1) / (2n+5)--------(i)

We need to find the ratio of their 5th Term.

a₅ = a+4d
a₅' = a'+4d'

Comparing From (i) with the a₅ for the value of n,

n-1/2 = 4
n-1 = 8
n = 9 --------(ii)

From (i)
Substituting the value on n,

[a+(n-1/2)d] / [a'+(n-1/2)d'] = (2n+1) / (2n+5)

[a+(9-1/2)d] / [a'+(9-1/2)d'] = (2(9)+1) / (2(9)+5)

[a+4d] / [a'+4d'] = (18+1) / (18+5)

[a+4d] / [a'+4d'] = 19 / 23

a₅ / a₅' = 19 / 23

∴, The ratio of their 5th term 19:23.

Hope it Helped U!!