Question

if the sum of the first 4terms of an Ap is 40 and that of first 14 terms is 280find the sum of the first n term
​

Answer 1

$\sf{ \sf S_n = n^2 + 6n}$

Given:-

• $\sf{Sum\;of\;first\;4\;terms\;of\;an\;AP\;( \sf S_4 ) = 40}$

• $\sf{Sum\;of\;first\;14\;terms\;of\;an\;AP\;( \sf S_{14} ) = 280}$

To Find:-

• $\sf{Sum\;of\;first\;n\;terms\;of\;an\;AP\;( \sf S_n ) = ?}$

Solution:-

$\small \; \; \; \star \; {\underline{\underline{\sf{\red{Sum\;of\;n\;terms\;of\;AP\;:}}}}}$

${\underline{\boxed{\sf{\dag\; \sf S_n = \dfrac{n}{2} \bigg( 2a + (n - 1)d \bigg)}}}}$

$\small \; \; \; \star \; {\underline{\underline{\sf{\red{Sum\;of\;4\;terms\;of\;AP\;:}}}}}$

$: \implies \sf{ \sf S_4 = \cancel{ \dfrac{4}{2}} \bigg( 2a + (4 - 1)d \bigg)}$

$: \implies \sf{40 = 2 \bigg( 2a + 3d \bigg)}$

$: \implies \sf{ \cancel{ \dfrac{40}{2}}= 2a + 3d}$

$: \implies \sf{20 = 2a + 3d}$ ---- eq.(1)

Similarly,

$\small \; \; \; \star \; {\underline{\underline{\sf{\red{Sum\;of\;14\;terms\;of\;AP\;:}}}}}$

$: \implies \sf{ \sf S_{14} = \cancel{ \dfrac{14}{2}} \bigg( 2a + (14 - 1)d \bigg)}$

$: \implies \sf{280 = 7 \bigg( 2a + 13d \bigg)}$

$: \implies \sf{ \cancel{ \dfrac{280}{7}} = 2a + 13d}$

$: \implies \sf{40 = 2a + 13d}$ ---- eq.(2)

$\rule{200}{2}$

$\small \; \; \; \star \; {\underline{\underline{\sf{Subtracting\;eq(2)\;from\;eq(1):}}}}$

we get,

$: \implies \sf{-10d = -20}$

$: \implies \sf{d = \cancel{ \dfrac{-20}{-10}}}$

$: \implies {\underline{\sf{\purple{d = 2}}}}$

$\small \; \; \; \star \; {\underline{\underline{\sf{Putting\;value\;of\;d\;in\;eq(1):}}}}$

$: \implies \sf{20 = 2a + 3 \times 2}$

$: \implies \sf{20 = 2a + 6}$

$: \implies \sf{20 - 6 = 2a}$

$: \implies \sf{14 = 2a}$

$: \implies \sf{a} = \cancel{ \dfrac{14}{2}}$

$: \implies {\underline{\sf{\purple{a = 7}}}}$

$\rule{200}{2}$

★ Now, we have value of a and d

$\small \; \; \; \star \; {\underline{\underline{\sf{Putting\;value\;of\;a\;and\;d\;in:}}}}$

${\underline{\boxed{\sf{\dag\; \sf S_n = \dfrac{n}{2} \bigg( 2a + (n - 1)d \bigg) }}}}$

$: \implies \sf{ \sf S_n = \dfrac{n}{2} \bigg( 2 \times 7 + (n - 1)2 \bigg) }$

$: \implies \sf{ \sf S_n = \dfrac{n}{2} \bigg( 14 + 2n - 2 \bigg) }$

$: \implies \sf{ \sf S_n = \dfrac{n}{2} \bigg( 2n + 12 \bigg) }$

$: \implies \sf{ \sf S_n = \dfrac{2n^2}{2} + \dfrac{12n}{2}}$

$: \implies \sf{ \sf S_n = \dfrac{ \cancel{2} n^2}{ \cancel{2}} + \dfrac{ \cancel{12} n}{ \cancel{2}}}$

$: \implies \sf{ \sf S_n = n^2 + 6n}$

${\underline{\underline{\sf{\purple{\dag \; Hence\;Solved!}}}}}$

$\rule{200}{2}$

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ sum \ of \ first \ n \ terms \ is \ n(n+6)}$

$\sf\orange{Given:}$

$\sf{In \ an \ AP,}$

$\sf{\implies{Sum \ of \ first \ 4 \ terms \ (S4)=40}}$

$\sf{\implies{Sum \ of \ first \ 14 \ terms \ (S14)=280}}$

$\sf\orange{To \ find:}$

$\sf{The \ sum \ of \ first \ n \ term \ (Sn).}$

$\sf\green{\underline{\underline{Solution:}}}$

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

$\sf{According \ to \ the \ first \ condition.}$

$\sf{S4=\frac{4}{2}[2a+(4-1)d]}$

$\sf{\therefore{40=2\times[2a+3d]}}$

$\sf{\therefore{2a+3d=\frac{40}{2}}}$

$\sf{\therefore{2a+3d=20...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{S14=\frac{14}{2}[2a+(14-1)d]}$

$\sf{\therefore{280=7\times[2a+6d]}}$

$\sf{\therefore{2a+13d=\frac{280}{7}}}$

$\sf{\therefore{2a+13d=40...(2)}}$

$\sf{Subtract \ equation (1) \ from \ equation (2)}$

$\sf{2a+13d=40}$

$\sf{-}$

$\sf{2a+3d=20}$

_____________________

$\sf{10d=20}$

$\boxed{\sf{\therefore{d=2}}}$

$\sf{Substitute \ d=2 \ in \ equation (1)}$

$\sf{2a+3\times2=20}$

$\sf{\therefore{2a+6=20}}$

$\sf{\therefore{2a=14}}$

$\sf{\therefore{a=\frac{14}{2}}}$

$\boxed{\sf{\therefore{a=7}}}$

_____________________________________

$\sf{Here, \ a=7 \ and \ d=2}$

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

$\sf{\therefore{Sn=\frac{n}{2}[2(7)+(n-1)\times2]}}$

$\sf{\therefore{Sn=n\times[7+(n-1)]}}$

$\sf{\therefore{Sn=n\times[n+6]}}$

$\sf{\therefore{Sn=n^{2}+6n}}$

$\sf{\therefore{Sn=n(n+6)}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ first \ n \ terms \ is \ n(n+6)}}}$