Advertisements
Advertisements
Question
If `(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ... "upto n terms")/(1 + 2 + 3 + 4 + ... "upto n terms") = 100/3,` find n
Advertisements
Solution
`(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ... "upto n terms")/(1 + 2 + 3 + 4 + ... "upto n terms") = 100/3,`
∴ `(sum_("r" = 1)^"n""r"("r" + 1))/(sum_("r" = 1)^"n" "r") = 100/3`
∴ `(sum_("r" = 1)^"n" "r"^2 + sum_("r" = 1)^"n" "r")/(sum_("r" = 1)^"n" "r") = 100/3`
∴ `(("n"("n" + 1)(2"n" + 1))/6 + ("n"("n" + 1))/2)/(("n"("n" + 1))/2) = 100/3`
∴ `(("n"("n" + 1))/6[(2"n" + 1) + 3])/(("n"("n" + 1))/2) = 100/3`
∴ `(2("n" + 2))/3= 100/3`
∴ n + 2 = 50
∴ n = 48
APPEARS IN
RELATED QUESTIONS
Find the sum to n terms 8 + 88 + 888 + 8888 + ...
Find the sum to n terms 0.4 + 0.44 + 0.444 + ...
Find the sum to n terms 0.7 + 0.77 + 0.777 + ...
Find Sn of the following arithmetico - geometric sequence:
1, 4x, 7x2, 10x3, 13x4, …
Find Sn of the following arithmetico - geometric sequence:
1, 2 × 3, 3 × 9, 4 × 27, 5 × 81, …
Find the sum to infinity of the following arithmetico - geometric sequence:
`1, 2/4, 3/16, 4/64, ...`
Find the sum to infinity of the following arithmetico - geometric sequence:
`1, -4/3, 7/9, -10/27 ...`
Find `sum_("r" = 1)^"n"(3"r"^2 - 2"r" + 1)`
Find `sum_("r" = 1)^"n"((1 + 2 + 3 .... + "r")/"r")`
Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ... upto n terms
Find the sum 22 + 42 + 62 + 82 + ... upto n terms
Find (702 – 692) + (682 – 672) + (662 – 652) + ... + (22 – 12)
If S1, S2 and S3 are the sums of first n natural numbers, their squares and their cubes respectively then show that - 9S22 = S3 (1 + 8 S1)
Answer the following:
Find `sum_("r" = 1)^"n" (5"r"^2 + 4"r" - 3)`
Answer the following:
Find `sum_("r" = 1)^"n" ((1^2 + 2^2 + 3^2 + ... + "r"^2)/(2"r" + 1))`
Answer the following:
Find `sum_("r" = 1)^"n" ((1^3 + 2^3 + 3^3 + ... "r"^3)/("r" + 1)^2)`
Answer the following:
Find 2 × 6 + 4 × 9 + 6 × 12 + ... upto n terms
Answer the following:
Find 2 × 5 × 8 + 4 × 7 × 10 + 6 × 9 × 12 + ... upto n terms
Answer the following:
Find `1^2/1 + (1^2 + 2^2)/2 + (1^2 + 2^2 + 3^2)/3 + ...` upto n terms
Answer the following:
Find 122 + 132 + 142 + 152 + ... 202
Answer the following:
Find (502 – 492) + (482 – 472) + (462 – 452) + ... + (22 – 12)
Answer the following:
If `(1 xx 3 + 2 xx 5 + 3 xx 7 + ... "upto n terms")/(1^3 + 2^3 + 3^3 + ... "upto n terms") = 5/9`, find the value of n
The sum of n terms of the series 22 + 42 + 62 + ........ is ______.
`(x + 1/x)^2 + (x^2 + 1/x^2)^2 + (x^3 + 1/x^3)^2` ....upto n terms is ______.
