Advertisements
Advertisements
Question
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
Advertisements
Solution
Given that,
a3 = 12
a50 = 106
We know that,
an = a + (n − 1)d
a3 = a + (3 − 1)d
12 = a + 2d ...(i)
Similarly, a50 = a + (50 − 1)d
106 = a + 49d ...(ii)
On subtracting (i) from (ii), we obtain
94 = 47d
d = 2
From equation (i), we obtain
12 = a + 2(2)
a = 12 − 4
a = 8
a29 = a + (29 − 1)d
a29 = 8 + (28)2
a29 = 8 + 56
a29 = 64
Therefore, 29th term is 64.
RELATED QUESTIONS
Ramkali required Rs 2,500 after 12 weeks to send her daughter to school. She saved Rs 100 in the first week and increased her weekly saving by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks.
What value is generated in the above situation?
Find the sum of first 8 multiples of 3.
If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.
Write an A.P. whose first term is a and common difference is d in the following.
a = 6, d = –3
There are 25 rows of seats in an auditorium. The first row is of 20 seats, the second of 22 seats, the third of 24 seats, and so on. How many chairs are there in the 21st row ?
If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then \[\frac{S_1}{S_2} =\]
If 18, a, b, −3 are in A.P., the a + b =
The first three terms of an A.P. respectively are 3y − 1, 3y + 5 and 5y + 1. Then, y equals
Q.2
Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
