Question

what is the sum of the first 10 term of an AP 41 ,36,31,26​

Answer 1

Answer:

Given: first term = a = 41 , common difference = d = 5

To Find: sum of first 10 terms (S10)

solution: we know, Sn=n/2(2a+(n-1) d)

hence, S10 = 10/2(2×41+(10-1) 5)

hence, S10 = 5×(82+45)

hence, S10 = 5×127

hence, S10 = 635

hence sum of first 10 terms of an AP is 635

Answer 2

Question :-

• What is the sum of the first 10 term of an AP 41, 36, 31, 26........?

Answer

Given :-

• An AP series 41, 36, 31, 26,........

To Find :-

• Sum of 10 terms.

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\bf \:S_n = \dfrac{n}{2} (2a + (n - 1)d)$

where,

• a = first term of an AP
• d = Common Difference of an AP

$\begin{gathered}\Large{\bold{\red{\underline{CaLcUlAtIoN\::}}}} \end{gathered}$

Here AP series,

$\bf \: ⟼ 41, 36, 31, 26,........$

$\begin{gathered}\bf\red{Let,} \end{gathered}$

$\begin{gathered}\longmapsto\:\:\bf{First \: term \:is \:a\: }\end{gathered}$

$\begin{gathered}\longmapsto\:\:\bf{Common \: Difference \:is \: d}. \\ \end{gathered}$

$\bf \: ⟼ a \: = 41$

$\bf \: ⟼ d = 36 - 41 = - 5$

$\bf \: ⟼ \bf \:S_n = \dfrac{n}{2} (2a + (n - 1)d)$

On substituting the values of a, d and n, we get

$\bf \: ⟼ S_{10} = \dfrac{10}{2} (2 \times 41 + (10 - 1) \times ( - 5))$

$\bf \: ⟼ S_{10} = 5 \times (82 - 45)$

$\bf \: ⟼ S_{10} = 5 \times 37$

$\bf \: ⟼ S_{10} = 185$

$\begin{gathered}\bf\red{So,}\end{gathered}$

$\bf \: ⟼ Sum \: of \: 10 \: terms \: = 185$