Advertisements
Advertisements
Question
The 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94. Find the common difference of the AP.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a9 = - 32
⇒ a +(9-1) d= -32 [an = a + (n-1) d]
⇒ a + 8d = -32 ............(1)
Now ,
`a_11 + a_13 = -94` (Given)
`⇒ ( a +10d ) +( a + 12d) = -94`
⇒ 2a + 22d = -94
⇒ a+ 11d = -47 ...............(2)
From (1) and (2), we get
-32 -8d + 11d=-47
⇒ 3d = -47 + 32 =-15
⇒ d = -5
Hence, the common difference of the AP is - 5.
APPEARS IN
RELATED QUESTIONS
If the mth term of an A.P. is 1/n and the nth term is 1/m, show that the sum of mn terms is (mn + 1)
If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Find the 12th term from the end of the following arithmetic progressions:
3, 5, 7, 9, ... 201
Find the sum of all natural numbers between 1 and 100, which are divisible by 3.
Mark the correct alternative in each of the following:
If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =
Q.3
Find S10 if a = 6 and d = 3
The nth term of an Arithmetic Progression (A.P.) is given by the relation Tn = 6(7 – n)..
Find:
- its first term and common difference
- sum of its first 25 terms
