Question

5th term of an arithmetic sequence is 30. 12th term is 65 then find the sum of 65 terms

Answer 1

Answer:

d=5

a65 =330

Sn = n/2(a+l)

s65=65/2( 10+330)

s65=65/2*340

=>65x170

s65= 10050

Answer 2

Answer :

1885

Explanation :

Here we have

$\sf a_5= 30\newline a+4d+30---(1)\newline \newline a_{12}=65 \newline a+11d=65\newline \newline Subtracting\:equation\:1\:from\:2\:we\:get \newline \newline {{a+11d=65} \atop {a+4d=30}}\newline -\rule{100}{1}\newline \newline7d=35 \newline \newline \longmapsto d=\dfrac{35}{7}=5$

$\sf Putting\:value\:of\:d\:in\:equation\:1\:we\:get \newline \newline \longmapsto a+4d=30\newline \newline \longmapsto a+4\times5=30\newline \newline \longmapsto a+20=30\newline \newline \longmapsto a=10 \newline \newline We\:know\:that \newline \newline s_n=\frac{n}{2}[2a+(n-1)d]\newline \newline\therefore s_{26}=\frac{26}{2}[2\times10+(26-1)5]\newline \newline\longmapsto s_{26}=13\times[20+25\times5]$

$\longmapsto s_{26}=13\times[20+125]\newline \newline\longmapsto s_{26}=13\times[145]\newline \newline\longmapsto s_{26}=1885$

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