Question

the sum of four consecutive term of an A. P. is 2.the sum of 3rd and 4th terms is 11.find the terms​

Answer 1

Answer:

Given:-

• The sum of four consecutive term of an AP is 2 .
• The sum of 3rd ans 4th terms is 11.

To find:-

All four terms

Solution:-

Let the terms are

• a+d,a+2d,a+3d,a+4d

ATQ,

$\\\qquad\quad\displaystyle\sf {:}\leadsto (a+d)+(a+2d)+(a+3d)+(a+4d)=2$

$\\\qquad\quad\displaystyle\sf {:}\leadsto a+d+a+2d+a+3d+a+4d=2$

$\\\qquad\quad\displaystyle\sf {:}\leadsto a+a+a+a+d+2d+3d+4d=2$

$\\\qquad\quad\displaystyle\sf {:}\leadsto 4a+10d=2$

$\\\qquad\quad\displaystyle\sf {:}\leadsto 2 (2a+5d=2$

$\\\qquad\quad\displaystyle\sf {:}\leadsto 2a+5d=\dfrac {2}{2}$

$\\\qquad\quad\displaystyle\sf {:}\leadsto 2a+5d=1 \dots\dots (1)$

Again,

$\\\qquad\quad\displaystyle\sf {:}\leadsto (a+3d)+(a+4d)=11$

$\\\qquad\quad\displaystyle\sf {:}\leadsto a+a+3d+4d=11$

$\\\qquad\quad\displaystyle\sf {:}\leadsto 2a+7d=11 \dots\dots (2)$

______________________________

By subtracting eq (2) from eq (1) we get

$\\\qquad\quad\displaystyle\sf {:}\leadsto -2d=-10$

$\\\qquad\quad\displaystyle\sf {:}\leadsto d=\dfrac {-10}{-2}$

$\\\qquad\quad\displaystyle\sf {:}\leadsto d=5$