Advertisements
Advertisements
Question
The first and the last terms of an AP are 8 and 65 respectively. If the sum of all its terms is 730, find its common difference.
Solution
Let a be the first term, d be the common difference and Tn be the last of the AP.
Given:
a = 8
Tn = 65
Sn = 730
We know:
Tn = a + (n − 1)d
⇒ 65 = 8 + (n − 1)d
⇒ 57 = (n − 1)d ...(i)
and `Sn=n/2(a+Tn)`
`730=n/2(8+65)`
`n/2=730/73`
n=2x10=20
On substituting n = 20 in (i), we get:
57 = (20 − 1)d
⇒ 57 = (19)d
⇒ d = 57/19=3
Thus, the common difference is 3.
APPEARS IN
RELATED QUESTIONS
How many two-digit number are divisible by 6?
The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms
Choose the correct alternative answer for the following question .
The sequence –10, –6, –2, 2,...
The sum of 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P.
Find the sum \[7 + 10\frac{1}{2} + 14 + . . . + 84\]
Which term of the sequence 114, 109, 104, ... is the first negative term?
Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.
Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th term is
The common difference of the A.P. \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\] is
If a = 6 and d = 10, then find S10