Question

The sum
of 4th and gith terms of an A.s. is
70, and the sum of 6th and 10th
sum of 6th and 10th term is 94.
Find the sum OF First 20 terms.​

Answer 1

Correct Question :-

The sum of 4th term and 8th term of an AP is 70, and the sum of 6th and 10th term is 94. Find the sum of first 20 terms.

Solution :-

Let :-

First term = a

Common difference = d

We know :-

$\bf a_n=a+(n-1)d$

$\bullet \ \sf a_4+a_8 = 70$

$\longrightarrow \sf [ \ a+(4-1)d \ ] +[ \ a+(8-1)d \ ] = 70 \\\\\longrightarrow a+3d+a+7d = 70 \\\\\longrightarrow 2a+10d = 70 \\\\\longrightarrow a+5d = 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .................(1)$

$\bullet \sf \ a_6+a_{10 }= 94$

$\longrightarrow \sf [ \ a+(6-1)d \ ] + [ \ a+(10-1)d \ ] = 94 \\\\\longrightarrow a+5d+a+9d = 94 \\\\\longrightarrow 2a+14d = 94 \\\\\longrightarrow a+7d = 47 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .................(2)$

Equation (2) - Equation (1) :-

$\longrightarrow \sf 2d= 12 \\\\\longrightarrow \bf d = 6$

Substitute the value of d in eq(1) :-

$\longrightarrow \sf a+ 5 \times 6 = 35 \\\\\longrightarrow a+30 = 35 \\\\\longrightarrow \bf a = 5$

$\underline{\boxed{\bf Sum \ of \ first \ n \ terms = \dfrac{n}{2} \times [ \ 2a+(n-1)d \ ] }}$

Sum of first 20 terms :-

$\longrightarrow \sf \dfrac{20}{2} \times [ \ 2 \times 5+(20-1) \times 6 \ ] \\\\\longrightarrow 10 \times [ \ 10+( 19 \times 6 ) \ ] \\\\\longrightarrow 10 \times [ \ 10+114 \ ] \\\\\longrightarrow 10 \times 124\\\\\longrightarrow \bf 1240$