Advertisements
Advertisements
प्रश्न
Find the sum:
`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms
उत्तर
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
संबंधित प्रश्न
Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
The sum of the first n terms in an AP is `( (3"n"^2)/2 +(5"n")/2)`. Find the nth term and the 25th term.
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.
There are 25 rows of seats in an auditorium. The first row is of 20 seats, the second of 22 seats, the third of 24 seats, and so on. How many chairs are there in the 21st row ?
Find where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .
If Sn denote the sum of n terms of an A.P. with first term a and common difference dsuch that \[\frac{Sx}{Skx}\] is independent of x, then
Q.16
Find the sum of all members from 50 to 250 which divisible by 6 and find t13.
If a = 6 and d = 10, then find S10
Solve the equation:
– 4 + (–1) + 2 + 5 + ... + x = 437
