Advertisements
Advertisements
प्रश्न
If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.
Advertisements
उत्तर
T3 = 1, T6 = –11, n = 32
a + 2d = 1 ...(i)
a + 5d = –11 ...(ii)
– – +
Subtracting (i) and (ii),
–3d = 12
⇒ d = `(12)/(-3)` = –4
Substitute the value of d in eq. (i)
a + 2(–4) = 1
⇒ a – 8 = 1
a = 1 + 8 = 9
∴ a = 9, d = –4
S32 = `n/(2)[2a + (n - 1)d]`
= `(32)/(2)[2 xx 9 + (32 - 1) xx (–4)]`
= 16[18 + 31 x (–4)]
= 16[18 – 124 ]
= 16 x (–106)
= –1696.
APPEARS IN
संबंधित प्रश्न
Find the sum of the following APs.
`1/15, 1/12, 1/10`, ......, to 11 terms.
Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.
Find the sum of the first n natural numbers.
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).
There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
Write 5th term from the end of the A.P. 3, 5, 7, 9, ..., 201.
Q.2
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.
If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is ______.
