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

$\underline\mathfrak{Given:-}$

• $\: \: \: \: \: The \: \: sum \: \: of \: \: 3rd \: \: and, \: \: 11th \: \: term's \: \: is \: \: {154}$

• $\: \: \: \: \: The \: \: sum \: \: of \: \: 6th \: \: and, \: \: 10th \: \: term's \: \: is \: \: {162}$

$\underline\mathfrak{To \: \: Find:-}$

• $\: \: \: \: \: find \: \: the \: \: sum \: \: of \: \: first \: \: 15th \: \: term.?$

$\huge\underline\mathfrak{Solutions:-}$

$\: \: \: \: \: \: \: \fbox{{A_n} \: \: = \: \: {a} \: + \: (n \: - \: 1) \: d}$

• $\: \: \: \: \: {3rd} \: \: = \: \: {a} \: + \: 2d$

• $\: \: \: \: \: {11th} \: \: = \: \: {a} \: + \: 10d$

• $\: \: \: \: \: {6th} \: \: = \: \: {a} \: + \: 5d$

• $\: \: \: \: \: {10th} \: \: = \: \: {a} \: + \: 9d$

$\: \: \: \: \: The \: \: sum \: \: of \: \: 3rd \: \: and, \: \: 11th \: \: term's \: \: is \: \: {154}$

$\: \: \: \: \: \leadsto \: \: {A_3} \: \: + \: \: {A_{11}} \: \: = \: \: {154}$

$\: \: \: \: \: \leadsto {(a \: + \: 2d)} \: + \: {(a \: + \: 10d)} \: \: = \: \: {154}$

$\: \: \: \: \: \leadsto {2a} \: + \: {12d} \: \: = \: \: {154}$

$\: \: \: \: \: \leadsto {2 \: ({a} \: + \: {6d})} \: \: = \: \: {154}$

$\: \: \: \: \: \leadsto {{a} \: + \: {6d}} \: \: = \: \: \frac{154}{2}$

$\: \: \: \: \: \leadsto {{a} \: + \: {6d}} \: \: = \: \: {77} \: …...(1).$

$\: \: \: \: \: The \: \: sum \: \: of \: \: 6th \: \: and, \: \: 10th \: \: term's \: \: is \: \: {162}$

$\: \: \: \: \: \leadsto \: \: {A_6} \: \: + \: \: {A_{10}} \: \: = \: \: {162}$

$\: \: \: \: \: \leadsto {(a \: + \: 5d)} \: + \: {(a \: + \: 9d)} \: \: = \: \: {162}$

$\: \: \: \: \: \leadsto {2a} \: + \: {14d} \: \: = \: \: {162}$

$\: \: \: \: \: \leadsto {2 \: ({a} \: + \: {7d})} \: \: = \: \: {162}$

$\: \: \: \: \: \leadsto {{a} \: + \: {7d}} \: \: = \: \: \frac{162}{2}$

$\: \: \: \: \: \leadsto {{a} \: + \: {7d}} \: \: = \: \: {81} \:..…...(2).$

$\: \: \: \: \: \underline{By \: \: Elimination \: \: Mathod}$

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

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

$\underline{(-) \: \: \: \: \: \: (-) \: = \: \: (-) \: \: }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {d} \: \: = \: \: {4}$

$\: \: \: \: \: Now, \: \: putting \: \: the \: \: value \: \: of \: \: d \: \: in \: \: Eq. \: {1}$

$\: \: \: \: \: \fbox{{{a} \: + \: {6d}} \: \: = \: \: {77}.}$

$\: \: \: \: \: \leadsto {a} \: + \: {{6} \: \times \: {4}} \: \: = \: \: {77}$

$\: \: \: \: \: \leadsto {{a} \: + \: {24}} \: \: = \: \: {77}$

$\: \: \: \: \: \leadsto {a} \: \: = \: \: {{77} \: - \: {24}}$

$\: \: \: \: \: \leadsto {a} \: \: = \: \: {53}$

$\: \: \: \: \: \underline{So, \: \: {a} \: \: = \: \: {53}, \: \: {d} \: \: = \: \: {4}, \: \: and \: \: {n} \: \: = \: \: {15}}$

$\: \: \: \: \: \: \: \fbox{{S_n} \: {\frac{15}{2} \: [2a \: + \: (n \: - \: 1) \: d]}}$

$\: \: \: \: \: \: \: \leadsto{{S_{15}} \: \: = \: \: \frac{15}{2} \: {[2\: {(15)} \: + \: (15 \: - \: 1) \: 4]}}$

$\: \: \: \: \: \: \: \leadsto{{S_{15}} \: \: = \: \: \frac{15}{2} \: {[{106} \: + \: {56}]}}$

$\: \: \: \: \: \: \: \leadsto{{S_{15}} \: \: = \: \: \frac{15}{2} \: \times \: {162}}$

$\: \: \: \: \: \: \: \leadsto{{S_{15}} \: \: = \: \: {15} \: \times \: {81}}$

$\: \: \: \: \: \: \: \leadsto{{S_{15}} \: \: = \: \: {1215}}$

$\: \: \: \: \: The \: \: sum \: \: of \: \: 1st \: \: {15} \: \: is \: \: {1215}.$

____________________________________

Answer 2

Answer:

The sum of 3

rd

and 11

th

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

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

th

and 10

th

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

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

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

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

Solution:

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

tn=a+(n−1)d

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

\sf{t3+t11=154}t3+t11=154

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

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

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

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

2

154

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

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

\sf{t6+t10=162}t6+t10=162

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

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

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

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

2

162

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

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

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

\sf{-}−

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

_________________

\boxed{\sf{d=4}}

d=4

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

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

\sf{\therefore{a+24=77}}∴a+24=77

\sf{\therefore{a=77-24}}∴a=77−24

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

∴a=53

____________________________________

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

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

Sn=

2

n

[2a+(n−1)d]

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

2

15

[2(53)+(15−1)4]

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

2

15

[106+56]

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

2

15

×162

\sf{\therefore{S15=15\times81}}∴S15=15×81

\sf{\therefore{S15=1215}}∴S15=1215

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