Advertisements
Advertisements
प्रश्न
In an A.P. the first term is 25, nth term is –17 and the sum of n terms is 132. Find n and the common difference.
उत्तर
First term a = 25
nth term = –17
`=>` Last term l = –17
Sum of n terms = 132
`=> n/2 [a + l] = 132`
`=>` n(25 – 17) = 264
`=>` n × 8 = 264
`=>` n = 33
Now, l = –17
`=>` a + (n – 1)d = –17
`=>` 25 + 32d = –17
`=>` 32d = – 42
`=> d = -42/32`
`=> d = -21/16`
APPEARS IN
संबंधित प्रश्न
The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`
Find the sum of all integers between 84 and 719, which are multiples of 5.
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
Find the sum of first n even natural numbers.
Choose the correct alternative answer for the following question .
If for any A.P. d = 5 then t18 – t13 = ....
The sum of the first n terms of an A.P. is 4n2 + 2n. Find the nth term of this A.P.
Write the common difference of an A.P. whose nth term is an = 3n + 7.
If `4/5` , a, 2 are three consecutive terms of an A.P., then find the value of a.
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
Find the sum of the integers between 100 and 200 that are
- divisible by 9
- not divisible by 9
[Hint (ii) : These numbers will be : Total numbers – Total numbers divisible by 9]