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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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Find the sum of first n odd natural numbers
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Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.
For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, ... and 3, 10, 17, ... equal?
Write the sum of first n odd natural numbers.
Q.5
Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
Assertion (A): a, b, c are in A.P. if and only if 2b = a + c.
Reason (R): The sum of first n odd natural numbers is n2.
