Question

in the ral
28. The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is
400. Find the first term and the common difference of the AP. [CBSE 2009)
the sum of its first 17​

Answer 1

Solution :-

Let :-

First term = a

Common difference = d

We know :-

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

$\bullet \sf \ S_9 = 81$

  $\longrightarrow \sf \dfrac{9}{2} \times [ \ 2a + (9-1)d \ ] = 81 \\\\\longrightarrow \dfrac{9}{2} \times [ \ 2a + 8d \ ] = 81 \\\\\longrightarrow 2a+8d = 81 \times \dfrac{2}{9} \\\\\longrightarrow 2a+8d = 18 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..................(1)$

$\bullet \sf \ S_{20} = 400$

  $\longrightarrow \sf \dfrac{20}{2} \times [ \ 2a + (20-1)d \ ] = 400 \\\\\longrightarrow 10 \times [ \ 2a+19d \ ] = 400 \\\\\longrightarrow 2a+19d= 40 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..................(2)$

Eq (2) - Eq (1)

  $\longrightarrow \sf 11d = 22 \\\\\longrightarrow \bf d = 2$

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

  $\longrightarrow \sf 2a+ 8 \times 2 = 18 \\\\\longrightarrow 2a + 16 = 18 \\\\\longrightarrow 2a= 2 \\\\\longrightarrow \bf a = 1$

Common difference = 2

First term = 1

Sum of first 17 terms :-

$\longrightarrow \sf \dfrac{17}{2} \times [ \ 2 \times1+(17-1)\times 2 \ ] \\\\\longrightarrow \dfrac{17}{2} \times [ \ 2+16 \times 2 \ ] \\\\\longrightarrow \dfrac{17}{2} \times [ \ 2+32 \ ] \\\\\longrightarrow \dfrac{17}{2} \times 34 \\\\\longrightarrow 17 \times 17 \\\\\longrightarrow \bf 289$