Let a1, a2, a3, .. . be terms of an A.P. If a1+a2+⋯+apa1+a2+⋯+aq=p2q2, p≠q, then a6a21 equals

Given, a1, a2, a3, ... are terms of A.P.∴ a1+a2+⋯+apa1+a2+⋯+aq=p2q2⇒ p22a1+(p−1)dq22a1+(q−1)d=p2q2 or  2a1+(p−1)d2a1+(q−1)d=pqor   2a1+(p−1)dq=2a1+(q−1)dpor   2a1(q−p)=d[(q−1)p−(p−1)q]or   2a1(q−p)=d(q−p)or   2a1=d∴  a6a21=a1+5da1+20d=a1+10a1a1+40a1=1141