Question

nth term of an AP isin an if a1+a2+a3=102 and a1=15 then find a10​

Answer 1

Step-by-step explanation:

a+a+d+a+2d = 102

3a + 3d = 102

a + d = 34

d = 34-a = 34-15 = 19

A(10) = a+9d = 15+9(19) = 15+171 = 186

Answer 2

Given:

${a = 15}$

$\sf{a_1+a_2+a_3=102--(i)}$

To find:

$\sf{a_{10}}$ = ?

Solution:

From (i)

$\sf{\implies a+a+d+a+2d=102}$

$\sf{\implies 3a+3d=102}$

$\sf{\implies a+d=34}$

put the value of a in the above equation.

$\sf{\implies 15+d=34}$

$\bf{\therefore d=19}$

_______________

$\sf{Now,\:a_{10}=a+9d}$

$\sf{\implies a_{10}=15+9(19)}$

$\sf{\implies a_{10}=15+171}$

$\boxed{\red{\bf{\therefore a_{10}=186}}}$

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