Advertisements
Advertisements
Question
Find a, b and c such that the following numbers are in AP: a, 7, b, 23, c.
Solution
For a, 7, b, 23, c... to be in AP
It has to satisfy the condition,
a5 – a4 = a4 – a3 = a3 – a2 = a2 – a1 = d
Where d is the common difference
7 – a = b – 7 = 23 – b = c – 23 ...(1)
Let us equation,
b – 7 = 23 – b
2b = 30
b = 15 ...(Equation 1)
And,
7 – a = b – 7
From equation 1
7 – a = 15 – 7
a = – 1
And,
c – 23 = 23 – b
c – 23 = 23 – 15
c – 23 = 8
c = 31
So a = – 1
b = 15
c = 31
Then, we can say that, the sequence – 1, 7, 15, 23, 31 is an AP.
APPEARS IN
RELATED QUESTIONS
Find the 60th term of the A.P. 8, 10, 12, ……., if it has a total of 60 terms and hence find the sum of its last 10 terms.
Find the middle term of the A.P. 213, 205, 197, ---, 37.
In the following situation, involved make an arithmetic progression? and why?
The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Following are APs or not? If they form an A.P. find the common difference d and write three more terms:
a, 2a, 3a, 4a …
For the following arithmetic progressions write the first term a and the common difference d:
−5, −1, 3, 7, ...
Which of the following sequences are arithmetic progressions? For those which are arithmetic progressions, find out the common difference.
12, 2, −8, −18, ...
Find 9th term of the A.P `3/4, 5/4, 7/4, 9/4,.....`
Find n if the given value of x is the nth term of the given A.P.
−1, −3, −5, −7, ...; x = −151
If (m + 1)th term of an A.P. is twice the (n + 1)th term, then prove that (3m + 1)th term is twice the (m + n + 1)th term
1, 6, 11, 16 ...... Find the 18th term of this A.P.
