Advertisements
Advertisements
Question
The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Solution
Let the first term, common difference and the number of terms of an AP are a, d and n respectively.
Given that, first term (a) = −5 and
Last term (l) = 45
Sum of the terms of the AP = 120
⇒ Sn = 120
We know that, if last term of an AP is known, then sum of n terms of an AP is,
Sn = `n/2 (a + 1)`
⇒ 120 = `n/2(-5 + 45)`
⇒ 120 × 2 = 40 × n
⇒ n = 3 × 2
⇒ n = 6
∴ Number of terms of an AP is known, then the nth term of an AP is,
l = a + (n – 1)d
⇒ 45 = –5 + (6 – 1)d
⇒ 50 = 5d
⇒ d = 10
So, the common difference is 10.
Hence, number of terms and the common difference of an AP are 6 and 10 respectively.
APPEARS IN
RELATED QUESTIONS
The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of numbers of houses preceding the house numbered x is equal to sum of the numbers of houses following x.
In an AP, given a = 2, d = 8, and Sn = 90, find n and an.
The first and last terms of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?
How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?
Find the sum of all odd numbers between 100 and 200.
Is 184 a term of the AP 3, 7, 11, 15, ….?
Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP.
Find the sum of first 17 terms of an AP whose 4th and 9th terms are –15 and –30 respectively.
If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
If the last term of an A.P. of 30 terms is 119 and the 8th term from the end (towards the first term) is 91, then find the common difference of the A.P. Hence, find the sum of all the terms of the A.P.