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प्रश्न
Find the sum of first 16 terms of the A.P. whose nth term is given by an = 5n – 3.
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उत्तर
Given, nth term of A.P. is an = 5n – 3
∴ a1 = 5(1) – 3 = 2
And a2 = 5(2) – 3 = 7
∴ Common difference, d = 7 – 2 = 5
∴ Sum of n terms of A.P., Sn = `n/2[2a + (n - 1)d]`
∴ S16 = `16/2 [2(2) + (16 - 1)5]`
= 8(4 + 75)
= 8 × 79
= 632
संबंधित प्रश्न
Find the sum given below:
–5 + (–8) + (–11) + ... + (–230)
In an A.P., if the first term is 22, the common difference is −4 and the sum to n terms is 64, find n.
If (2p +1), 13, (5p -3) are in AP, find the value of p.
Write an A.P. whose first term is a and common difference is d in the following.
a = 6, d = –3
If the sums of n terms of two arithmetic progressions are in the ratio \[\frac{3n + 5}{5n - 7}\] , then their nth terms are in the ratio
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
If the sum of first n terms of an AP is An + Bn² where A and B are constants. The common difference of AP will be ______.
The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)
Find the sum of first 20 terms of an A.P. whose nth term is given as an = 5 – 2n.
