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Question
The 13th terms of an AP is 4 times its 3rd term. If its 5th term is 16, Find the sum of its first 10 terms.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a13 = 4 × a3 (Given)
⇒ a +12d = 4 (a + 2d) [ an = a + (n-1) d]
⇒ a +12d = 4a +8d
⇒ 3a = 4d .................(1)
Also,
a5 = 16 (Given)
⇒ a +4d = 16 ..........(2)
Solving (1) and (2), we get
a +3a = 16
⇒ 4a = 16
⇒ a =4
Putting a = 4 in (1), we get
4d = 3× 4 = 12
⇒ d = 3
`"Using the formula ", S_4 = n/2 [ 2a + (n-1) d ] `, we get
`S_10 = 10/2 [2xx4+(10-1) xx3]`
=5 × ( 8 + 27)
= 5× 35
=175
Hence, the required sum is 175.
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