If the ratio of the 11th term of an AP to its 18th term is 2 : 3, find the ratio of the sum of the first five terms to the sum of its first 10 terms.
Given:
T11: T18 = 2: 3
We know that,
nth term = Tn = a + (n – 1)d
where,
a = first term
d = common difference
∴
T11 = a + (11 – 1)d
⇒ T11 = a + 10d …(1)
And,
T18 = a + (18 – 1)d
⇒ T18 = a + 17d …(2)
From (1) and (2),
![]()
⇒ 3(a + 10d) = 2(a + 17d)
⇒ 3a + 30d = 2a + 34d
⇒ 3a – 2a = 34d – 30d
⇒ a = 4d …(3)
We know that,
![]()
Now,

![]()
From (3),
![]()
![]()
![]()
![]()
![]()
Hence, S5:S10 = 6:17
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.

