Question

if the sum of 3rd and 11th term is 154 and sum of 6th ad 10th term is 162 find the sum of first 15th term​

Answer 1

$\large\bf\underline{Given:-}$

• Sum of 3rd and 11th term = 154
• sum of 6th and 10th term = 162

$\large\bf\underline {To \: find:-}$

• Sum of first 15 terms.

$\huge\bf\underline{Solution:-}$

$\large \bf \: T_n = a +(n-1)d$

sum of 3rd and 11th term is 154

$\rm \longmapsto \: T_3 + T_{11}=154............(i) \\\\ \rm \longmapsto \: T_3 = a + 2d..........(ii)\\ \\\rm \longmapsto \: T_{11} = a + 10d...........(iii) \\\\ \rm \: putting \: value \: of \:T_3 \: and \: T_{11} \: in \: eq.(ii) \: and \: (iii)\\\\ \rm \longmapsto \: (a + 2d) + (a + 10d) = 154\\ \\\rm \longmapsto \:2a + 12d = 154 \\ \\\rm \text{Dividing both side by 2} \\\\ \rm \longmapsto \:a + 6d = 77.........(iv)$

And sum of 6th and 10th term is 162

$\rm \longmapsto\: T_6 + T_{10} = 162 \\\\ \rm \longmapsto\:T_6 = a + 5d ..........(v)\\ \\\rm \longmapsto\:T_{10} = a + 9d...........(vi) \\ \\\rm \: putting \: value \: of \:T_6 \: and \: T_{10} \: in \: eq.(v) \: and \: (vi) \\ \\\rm \longmapsto\:(a + 5d) + (a + 9d) = 162 \\ \\\rm \longmapsto\:2a + 14d = 162 \\ \\ \rm \text{Dividing both side by 2} \\ \\\rm \longmapsto\:a + 7d = 81.......(vii)$

Solving eq.(iv) and (vii) .

$\rm \: a + 6d = 77 \\ \rm \: a + 7d = 81 \\ \: - \: \: \: \: - \: \: \: \: - \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \: \: \: \: \: \: \rm - d \: = - 4 \\ \bf \: \: \: \: \: \: \: \: d = 4$

Substituting value of d = 4 in eq. (iv)

$\rm \longmapsto \: a + 6d = 77 \\ \\ \rm \longmapsto \: a + 6 \times 4 = 77 \\ \\ \rm \longmapsto \: a + 24 = 77 \\\\ \rm \longmapsto \: a = 77 - 24 \\\\ \bf\longmapsto \: a = 53$

Now, sum of first 15 terms.

$\large \bf \: S_n = \frac{n}{2} [2a + (n - 1)d]$

$\leadsto \rm \: S_{15} = \frac{15}{2} [2 \times 53 + 14 \times 4] \\\\ \leadsto \rm \: S_{15} = \frac{15}{2} [106 + 56] \\ \\\leadsto \rm \: S_{15} = \frac{15}{ \cancel2} [ \cancel{162}] \\\\ \leadsto \rm \: S_{15} = 15 \times 81 \\\\ \leadsto \bf \: S_{15} = 1215$

So,

Sum of 1st 15 terms is 1215.

Answer 2

Question:

In an AP if the sum of 3rd and 11th term is 154 and sum of 6th ad 10th term is 162. Find the sum of first 15th term.

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

$\sf{The \ sum \ of \ first \ 15 \ terms \ is \ 1215.}$

$\sf\orange{Given:}$

$\sf{\implies{The \ sum \ of \ 3^{rd} \ and \ 11^{th}}}$

$\sf{term \ is \ 154.}$

$\sf{\implies{The \ sum \ of \ 6^{th} \ and \ 10^{th}}}$

$\sf{term \ is \ 162.}$

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

$\sf{Sum \ of \ first \ 15 \ terms.}$

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

$\boxed{\sf{tn=a+(n-1)d}}$

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

$\sf{t3+t11=154}$

$\sf{\therefore{(a+2d)+(a+10d)=154}}$

$\sf{\therefore{2a+12d=154}}$

$\sf{\therefore{2(a+6d)=154}}$

$\sf{\therefore{a+6d=\frac{154}{2}}}$

$\sf{\therefore{a+6d=77...(1)}}$

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

$\sf{t6+t10=162}$

$\sf{\therefore{(a+5d)+(a+9d)=162}}$

$\sf{\therefore{2a+14d=162}}$

$\sf{\therefore{2(a+7d)=162}}$

$\sf{\therefore{a+7d=\frac{162}{2}}}$

$\sf{\therefore{a+7d=81...(2)}}$

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

$\sf{a+7d=81}$

$\sf{-}$

$\sf{a+6d=77}$

_________________

$\boxed{\sf{d=4}}$

$\sf{Substitute \ d=4, \ in \ equation (1)}$

$\sf{a+6(4)=77}$

$\sf{\therefore{a+24=77}}$

$\sf{\therefore{a=77-24}}$

$\boxed{\sf{\therefore{a=53}}}$

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$\sf{Here, \ a=53, \ d=4 \ and \ n=15}$

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

$\sf{\therefore{S15=\frac{15}{2}[2(53)+(15-1)4]}}$

$\sf{\therefore{S15=\frac{15}{2}[106+56]}}$

$\sf{\therefore{S15=\frac{15}{2}\times162}}$

$\sf{\therefore{S15=15\times81}}$

$\sf{\therefore{S15=1215}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ first \ 15 \ terms \ is \ 1215.}}}$