Advertisements
Advertisements
प्रश्न
The 5th term and the 9th term of an Arithmetic Progression are 4 and – 12 respectively.
Find:
- the first term
- common difference
- sum of 16 terms of the AP.
Advertisements
उत्तर
Let a and d be the first term and common difference of A.P.
Then by Tn = a + (n – 1)d
Given T5 = 4
`\implies` a + 4d = 4 ...(1)
And T9 = – 12
`\implies` a + 8d = – 12 ...(2)
Solving equations (1) and (2), we get
– 4d = 16
`\implies` d = – 4
Put this value in equation (1)
a + 4 × (– 4) = 4
a – 16 = 4
a = 20
∴ a. First term a = 20
b. Common difference d = – 4
c. Sum of n terms = `n/2 [2a + (n - 1)d`
∴ Sum of 16 terms = `16/2 [2 xx 20 + 15 xx (-4)]`
= 8 [40 – 60]
= 8 × (– 20)
= – 160
APPEARS IN
संबंधित प्रश्न
Find how many integers between 200 and 500 are divisible by 8.
Find the middle term of the AP 6, 13, 20, ..., 216.
Determine k so that (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.
If 18, a, (b – 3) are in AP, then find the value of (2a – b).
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.
Find the sum of all multiples of 9 lying between 300 and 700.
The Sum of first five multiples of 3 is ______.
The sum of first n terms of an A.P is 5n2 + 3n. If its mth terms is 168, find the value of m. Also, find the 20th term of this A.P.
Write 5th term from the end of the A.P. 3, 5, 7, 9, ..., 201.
In a Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find that:
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.
