The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16, find the sum of its first 10 terms.
The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16, find the sum of its first 10 terms.
Let a be the first term and d be the common difference of the AP. Then,
$a_{13}=4 \times a_{3}$ (Given)
$\Rightarrow a+12 d=4(a+2 d) \quad\left[a_{n}=a+(n-1) d\right]$
$\Rightarrow a+12 d=4 a+8 d$
$\Rightarrow 3 a=4 d \quad \ldots(1)$
Also,
$a_{5}=16$ (Given)
$\Rightarrow a+4 d=16 \quad \ldots$ (2)
Solving (1) and (2), we get
$a+3 a=16$
$\Rightarrow 4 a=16$
$\Rightarrow a=4$
Putting a = 4 in (1), we get
$4 d=3 \times 4=12$
$\Rightarrow d=3$
Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get
$S_{10}=\frac{10}{2}[2 \times 4+(10-1) \times 3]$
$=5 \times(8+27)$
$=5 \times 35$
$=175$
Hence, the required sum is 175 .
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