Advertisements
Advertisements
प्रश्न
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
Advertisements
उत्तर
Let 'a' be the first term and 'd' be the common difference of the given A.P.
Given,
S14 = 1050
`\implies 14/2 [2a + (14 - 1)d] = 1050`
`\implies` 7[2a + 13d] = 1050
`\implies` 2a + 13d = 150
`\implies` a + 6.5d = 75 ...(i)
And t14 = 140
`\implies` a + 13d = 140 ...(ii)
Subtracting (i) from (ii), we get
6.5d = 65
⇒ d = 10
⇒ a + 13(10) = 140
⇒ a = 10
Thus, 20th term = t20
= 10 + 19d
= 10 + 19(10)
= 200
APPEARS IN
संबंधित प्रश्न
The first and the last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.
If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is, S1)? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th, and the nth terms.
In an A.P., the sum of first n terms is `(3n^2)/2 + 13/2 n`. Find its 25th term.
A sum of ₹2800 is to be used to award four prizes. If each prize after the first is ₹200 less than the preceding prize, find the value of each of the prizes
Write an A.P. whose first term is a and common difference is d in the following.
a = –19, d = –4
In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
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.
If m times the mth term of an A.P. is eqaul to n times nth term then show that the (m + n)th term of the A.P. is zero.
Find where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .
Find the sum of all even numbers from 1 to 250.
