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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.
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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
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संबंधित प्रश्न
How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?
Which term of the A.P. `20, 19 1/4, 18 1/2, 17 3/4,` ..... is the first negative term?
First term and the common differences of an A.P. are 6 and 3 respectively; find S27.
Solution: First term = a = 6, common difference = d = 3, S27 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]` - Formula
Sn = `27/2 [12 + (27 - 1)square]`
= `27/2 xx square`
= 27 × 45
S27 = `square`
The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.
Q.11
The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.
For an A.P., if t1 = 1 and tn = 149, then find Sn.
Activitry :- Here t1= 1, tn = 149, Sn = ?
Sn = `n/2 (square + square)`
= `n/2 xx square`
= `square` n, where n = 75
The sum of all two digit odd numbers is ______.
In an AP, if Sn = n(4n + 1), find the AP.
In an A.P., if Sn = 3n2 + 5n and ak = 164, find the value of k.
