For the A.P. -3,-7,-11..., can we find `a_(30)-a_(20)` without actually finding `a_(30)` and `a_(20)`

For the AP -3, -7, -11, … can we find directly a_(30)-a_(20) without actually finding a_(30) and a_(20) ? Give reason for your answer.

For the AP -3, -7, -11, … can we find directly a_(30)-a_(20) without actually finding a_(30) and a_(20) ? Give reason for your answer.

For the AP -3, -7, -11, … can we find directly a_(30)-a_(20) without actually finding a_(30) and a_(20) ? Give reason for your answer.

In the AP -9 , -14, -19, -24, ……… a_(30) - a_(20) = ………….

Find the indiated terms in each of the following A.P. (i) 1, 7, 13, 19 ,…, 301 , a_(10) , a_(20) (ii) a= 22 , d = - 3, a_(n) , a_(30)

Find the indiated terms in each of the following A.P. (i) 1, 7, 13, 19 ,…, 301 , a_(10) , a_(20) (ii) a= 22 , d = - 3, a_(n) , a_(30)

Find the sum of first 24 terms of the A.P. a_(1) , a_(2), a_(3) ...., if it is know that a_(1) + a_(5) + a_(10) + a_(15) + a_(20) + a_(24) = 225

Find the sum of first 24 terms of the A.P. a_(1) , a_(2), a_(3) ...., if it is know that a_(1) + a_(5) + a_(10) + a_(15) + a_(20) + a_(24) = 225

If a_(n) = (n - 3)/4 , then find a_(11),a_(15) and hence find a_(15)/a_(11) .

n^(th) term of an A.P. is a_(n) . If a_(1)+a_(2)+a_(3) = 102 and a_(1) = 15 , then find a_(10) .