Advertisements
Advertisements
प्रश्न
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Advertisements
उत्तर
Given, t2 = 14 and t3 = 18
`=>` d = t3 – t2
= 18 – 14
= 4
Now t2 = 14
`=>` a + d = 14
`=>` a = 14 – 4
`=>` a = 10
Sum of n terms of an A.P. = `n/2[2a + (n - 1)d]`
∴ Sum of first 51 terms of an A.P. = `51/2 [2 xx 10 + 50 xx 4]`
= `51/2 [20 + 50 xx 4]`
= `51/2 [20 + 200]`
= `51/2 xx 220`
= 51 × 110
= 5610
APPEARS IN
संबंधित प्रश्न
If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.
The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
In an A.P., the sum of first n terms is `(3n^2)/2 + 13/2 n`. Find its 25th term.
Find the sum of the first n natural numbers.
Find out the sum of all natural numbers between 1 and 145 which are divisible by 4.
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
Let there be an A.P. with first term 'a', common difference 'd'. If an denotes in nth term and Sn the sum of first n terms, find.
Find S10 if a = 6 and d = 3.
Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to `((a + c)(b + c - 2a))/(2(b - a))`
