Question

If the 10th term of an AP is 47 and its first term is 2,find the sum of its first 15 terms.

Answer 1

$T_{10} = 47 = 2 + (10-1) d \\ \\ 47 = 2+9d \\ \\ d = 5 \\ \\ S_{n} = a*n+\frac{d*n(n-1)}{2} \\ \\ S_{15} = 2*15+\frac{5*15*14}{2} = 30+525 = 555$

Answer 2

Answer:

Sum of first 15 terms in given A.P = 555

Step-by-step explanation:

Let a , d are first term and common difference of an A.P

a = 2 (given )

/* we know that,

$\boxed {n^{th}\: term \:(a_{n})=a+(n-1)d}$ */

i ) Now ,

$Given\: a_{10}=47$

$\implies a+9d=47$

$\implies 2+9d=47$

$\implies 9d = 47-2$

$\implies d = \frac{45}{9}$

$\implies d = 5$ ---(1)

ii ) Sum of first n terms in A.P

(Sn) = $\frac{n}{2}[2a+(n-1)d]$

Here, substitute n = 15, a = 2, d = 5, we get

$S_{15} = \frac{15}{2}[2\times 2+(15-1)\times 5]$

=$\frac{15}{2}\times (4+70)$

= $\frac{15}{2}\times 74$

= $ 15 \times 37$

= $ 555 $

Therefore,

Sum of first 15 terms in given A.P = 555

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