Advertisements
Advertisements
Question
Find the sum:
`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms
Solution
Here, first term (A) = `(a - b)/(a + b)`
And common difference,
D = `(3a - 2b)/(a + b) - (a - b)/(a + b)`
= `(2a - b)/(a + b)`
∵ Sum of n terms of an AP,
Sn = `n/2[2a + (n - 1)d]`
⇒ Sn = `n/2{2((a - b))/((a + b)) + (n - 1) ((2a - b))/((a + b))}`
= `n/2{(2a - 2b + 2an - 2a - bn + b)/(a + b)}`
= `n/2((2an - bn - b)/(a + b))`
∴ S11 = `11/2{(2a(11) - b(11) - b)/(a + b)}`
= `11/2((22a - 12b)/(a + b))`
= `(11(11a - 6b))/(a + b)`
APPEARS IN
RELATED QUESTIONS
Find the sum of first 15 multiples of 8.
Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.
If the numbers a, 9, b, 25 from an AP, find a and b.
Write the next term of the AP `sqrt(2) , sqrt(8) , sqrt(18),.........`
The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34. Find its common difference.
The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is `(1)/(3)`. Calculate the first and the thirteenth term.
If the nth term of an AP is (2n +1), then the sum of its first three terms is ______.
The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Assertion (A): a, b, c are in A.P. if and only if 2b = a + c.
Reason (R): The sum of first n odd natural numbers is n2.
The sum of n terms of an A.P. is 3n2. The second term of this A.P. is ______.