Question

Given p A.P's, each of which consists of n terms. If their first terms are 1, 2,
3,......, p and common differences are 1, 3, 5, ...., 2p – 1 respectively, then
sum of the terms of all the progressions is
(A) = n (np+1)
(B) = n(p+1)
(C) n (n+1)
(D) none of these
​

Answer 1

Answer:

gggggggggggggggggggggggggggggggggggggģggggggggggggggggggggggggg

c

Answer 2

Answer:

np/2[np+1].

Step-by-step explanation:

Summation for 1, S1 , in which starting a1=1 and difference d =1.

So, S1 = n/2[2*1 + (n-1)1].

S2=n/2[2*2 + (n-1)3]

and so on it will go till the pth term.

Sp, where difference d=(2p-1) and a1=p.

Sp =n/2[2p + (n-1)(2p-1)].

So, S1 + S2 +...+Sp = n/2[2*1+(n-1)1] + n/2[2*2 + (n-1)3] +...+ n/2[2p + (n-1)(2p-1)].

=n/2[2*1 + (n-1)*1 + 2*2 + (n-1)3 + ...+ 2p + (n-1)(2p-1)].

=n/2[2*1 + 2*2 + 2*3 + ...2p + (n-1)(1+3+5+...+(2p-1))].

=n/2[2(1+2+3+..+p) + (n-1)(1+3+5+...+(2p-1))].

Since, we know that summation for odd terms 1 + 3 + 5 + 7 ..(2n-1) = n^2 and for 1+2+3+...+n is n(n-1)/2.

So, n/2[2*p(p+1)/2 + (n-1)p^2].

Which on solving we will get that.

np/2[p+1 + (n-1)p].

So, S1+S2+...Sp = np/2[np+1].