Advertisements
Advertisements
Question
The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.
Advertisements
Solution
Let the first three terms of the AP be (a-d) , a and (a+d) . then
(a-d) + a+(a+d) = 48
⇒ 3a = 48
⇒ a= 16
Now,
(a-d)× a = 4 (a+d) + 12 (Given)
⇒ (16-d) × 16 = 4(16 +d) +12
⇒ 256-16d = 64 +4d +12
⇒16d + 4d=256-76
⇒ 20d=180
⇒ d=9
When a = 16 and d = 9 ,
a-d = 16-9=7
a+d = 16+9=25
Hence, the first three terms of the AP are 7, 16, and 25.
APPEARS IN
RELATED QUESTIONS
Find the sum of first 30 terms of an A.P. whose second term is 2 and seventh term is 22
Find the sum given below:
`7 + 10 1/2 + 14 + ... + 84`
In an AP given an = 4, d = 2, Sn = −14, find n and a.
If the 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms?
The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.
The sum of the first n terms in an AP is `( (3"n"^2)/2 +(5"n")/2)`. Find the nth term and the 25th term.
Simplify `sqrt(50)`
Find the sum (−5) + (−8)+ (−11) + ... + (−230) .
Write the expression of the common difference of an A.P. whose first term is a and nth term is b.
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
Q.6
Q.2
Q.10
Find the sum of numbers between 1 to 140, divisible by 4
Find the sum of odd natural numbers from 1 to 101
If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are ______.
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
Find the sum:
`4 - 1/n + 4 - 2/n + 4 - 3/n + ...` upto n terms
Find the sum of the integers between 100 and 200 that are not divisible by 9.
Find the middle term of the AP. 95, 86, 77, ........, – 247.
