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In an AP: Given a = 5, d = 3, an = 50, find n and Sn.
Let there be an A.P. with the first term ‘a’, common difference’. If a denotes its nth term and Sn the sum of first n terms, find:
n and Sn, if a = 5, d = 3 and an = 50.
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Solution 1
Given that, a = 5, d = 3, an = 50
As an = a + (n − 1)d,
⇒ 50 = 5 + (n - 1) × 3
⇒ 3(n - 1) = 45
⇒ n - 1 = 15
⇒ n = 16
Now, Sn = `n/2 (a + a_n)`
Sn = `16/2 (5 + 50)`
Sn = 440
Solution 2
Here, we have an A.P. whose nth term (an), first term (a) and common difference (d) are given. We need to find the number of terms (n) and the sum of first n terms (Sn).
Here,
First term (a) = 5
Last term (`a_n`) = 50
Common difference (d) = 3
So here we will find the value of n using the formula, `a_n = a + (n -1)d`
So, substituting the values in the above-mentioned formula
50 = 5 + (n -1)3
50 = 5 = 3n - 3
50 = 2 + 3n
3n = 50 - 2
Further simplifying for n
3n = 48
n = `48/3`
n = 16
Now, here we can find the sum of the n terms of the given A.P., using the formula,
Sn = `(n/2)(a + 1)`
Where a is the first term
l = the last term
So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,
S16 = `(16/2) [5 + 50]`
= 8(55)
= 440
Therefore, for the given A.P n = 16 and S16 = 440
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