Advertisements
Advertisements
प्रश्न
Find the sum of all 11 terms of an A.P. whose 6th term is 30.
Advertisements
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
If Sn1 denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Determine the nth term of the AP whose 7th term is -1 and 16th term is 17.
Write an A.P. whose first term is a and common difference is d in the following.
a = –3, d = 0
Choose the correct alternative answer for the following question .
In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....
If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is
Q.15
In a Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find that:
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
Find the value of x, when in the A.P. given below 2 + 6 + 10 + ... + x = 1800.
