For the series, `S=1+1/((1+3))(1+2)^2+1/((1+3+5))(1+2+3)^2+1/((1+3+5+7))(1+2+3+4)^2` +... 7th term is 16 7th term is 18 Sum of first 10 terms is `(505)/4` Sum of first 10 terms is `(45)/4`

For the series, S=1+1/((1+3))(1+2)^2+1/((1+3+5))(1+2+3)^2+1/((1+3+5+7))(1+2+3+4)^2 +... a.7th term is 16 b.7th term is 18 c. Sum of first 10 terms is (505)/4 d. Sum of first 10 terms is (45)/4

For the series, S=1+1/((1+3))(1+2)^2+1/((1+3+5))(1+2+3)^2+1/((1+3+5+7))(1+2+3+4)^2 +... (a) 7th term is 16 (b) 7th term is 18 (c) Sum of first 10 terms is (505)/4 (d) Sum of first 10 terms is (45)/4

For the series S=1+(1)/((1+3))(1+2)^(2)+(1)/((1+3+5))(1+2+3)^(2)+(1)/((1+3+5+7))(1+2+3+4)^(2)+……….., if the sum of the first 10 terms is K, then (4K)/(101) is equal to

For the series S=1+(1)/((1+3))(1+2)^(2)+(1)/((1+3+5))(1+2+3)^(2)+(1)/((1+3+5+7))(1+2+3+4)^(2)+…………….., if the 7^("th") term is K, then (K)/(4) is equal to

For the series S=1+(1)/((1+3))(1+2)^(2)+(1)/((1+3+5))(1+2+3)^(2)+(1)/((1+3+5+7))(1+2_3+4)^(2)+…………….., if the 7^("th") term is K, then (K)/(4) is equal to

In an A.P., if 5 ^(th) term is (1)/(7) and 7^(th) term is (1)/(5), then the sum of first 35 terms is

If the fourth term of a H.P is 1/3 and 7th term is 1/4, then 16th term is

Find the sum of first n terms and the sum of first 5 terms of the geometric series 1+2/3+4/9+......

If the fourth term of a H.P is 1/3 and 7 th term is 1/4 ,then 16 th term is

FIn dthe sum of this series (1^(3))/(1)+(1^(3)+2^(3))/(1+3)+(1^(3)+2^(3)+3^(3))/(1+3+5)+...+ till 10^(th) term ^(*)