Question

18th question plzz guys ty in advance..

Answer 1

(1)given:-
S₇=49
S₁₇=289
∵S₇=49
7{2a+(7-1)d}/2=49[∵Sₙ=n{2a+(n-1)d}/2]
⇒(2a+6d)/2 =7
⇒2(a+3d)/2=7
⇒a=7-3d.......(1)
also∵S₁₇=289
∴17{2a+(17-1)d}/2=289[∵Sₙ =n{2a+(n-1)d}/2]
⇒(2a+16d)/2=17
⇒2(a+8d)/2=17
⇒a=17-8d.........(2)
equating(1)&(2)
⇒7-3d=17-8d
⇒5d=10
⇒d=2
∴a=7-3(2)[using(1)]
a=1
∴Sₙ=n{2a+(n-1)d}/2
⇒Sₙ=n{2(1)+2(n-1)}/2
⇒Sₙ=n×2(1+n-1)/2
⇒Sₙ=n²
⇒(2)
given
a=12
d=6
aₗ=252
as per condition
aₗ=a+(l-1)d[∵aₙ=a+(n-1)d]
∴252=12+6(l-1)
⇒6(42)=6{2+(l-1)}
⇒42=2+l-1
⇒l=41
Sₗ=l(a+aₗ)/2
⇒Sₗ=41(12+252)/2
⇒Sₗ=41(264)/2
⇒Sₗ=5412