Question

the 5th term & 12th term of an
Arithematic progression are
30 & 65 respectively.
Find the sum of its 26th term .​

Answer 1

Answer:

given,

a5=30

a12=65

s26=?

a5=a+(n-1)d

30=a+(5-1)d

30=a+4d

30-4d=a --------(i)

a12=a+(n-1)d

substituting the value of a from (i),

65=30-4d+(12-1)d

65=30-4d+11d

65=30+7d

65-30=7d

35/7=d

5=d

substituting the value of d in (i)

30-4(5)=a

30-20=a

10=a

s26=n/2[2a+(n-1)d]

=26/2[2×10+(26-1)5]

=13(20+125)

=13×145

=1885

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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