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Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Posted by Khushi Yadav 6 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Sia ? 6 years, 3 months ago

The general term of an AP is given by a_n=a+(n-1)d and S_n={tex}\frac{n}{2}{/tex}[2a+(n-1)d].
Given that a₂=14 and a₃=18
So, d=a₃-a₂=18-14=4
Now, a₂=14 {tex}\Rightarrow{/tex}a+4=14 {tex}\Rightarrow{/tex}a=10
Also, S₅₁={tex}\frac{{51}}{2}{/tex}[2(10)+(50)4]
{tex}\Rightarrow{/tex}S₅₁={tex}\frac{{51}}{2}{/tex}[20+200]
{tex}\Rightarrow{/tex}S₅₁={tex}\frac{{51}}{2}{/tex}[220]
{tex}\Rightarrow{/tex}S₅₁=51{tex}\times{/tex}110
{tex}\Rightarrow{/tex}S₅₁=5610

1Thank You

Soyab Khan 6 years, 3 months ago

Tayyaua

0Thank You

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CBSE > Class 10 > Mathematics