Advertisements
Advertisements
प्रश्न
Find the sum of all 11 terms of an A.P. whose 6th term is 30.
उत्तर
Given, 6th term of A.P. = 30
or, a6 = 30
or, a + (6 – 1)d = 30
or, a + 5d = 30 ...(i)
Since, Sum of n terms of A.P. is Sn = `n/2[2a + (n - 1)d]`
∴ S11 = `11/2[2a + (11 - 1)d]`
= `11/2[2d + 10d]`
= `(11 xx 2)/2 [a + 5d]`
= 11 × 30 ...[From equation (i)]
= 330
APPEARS IN
संबंधित प्रश्न
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Find the sum of the following arithmetic progressions:
a + b, a − b, a − 3b, ... to 22 terms
Find the sum of the first 40 positive integers divisible by 3
How many terms of the A.P. : 24, 21, 18, ................ must be taken so that their sum is 78?
The 7th term of the an AP is -4 and its 13th term is -16. Find the AP.
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d).
The sum of third and seventh term of an A. P. is 6 and their product is 8. Find the first term and the common difference of the A. P.
Suppose three parts of 207 are (a − d), a , (a + d) such that , (a + d) >a > (a − d).
Q.10