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प्रश्न
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
उत्तर १
Given that,
a2 = 14
a3 = 18
d = a3 − a2
= 18 − 14
= 4
a2 = a + d
14 = a + 4
a = 10
Sn = `n/2[2a + (n - 1)d]`
S51 = `51/2[2 xx 10 + (51 - 1)4]`
=`51/2[20 + (50)(4)]`
=`(51(220))/2`
= 51 × 110
= 5610
उत्तर २
In the given problem, let us take the first term as a and the common difference as d
Here, we are given that,
a2 = 14 ...(1)
a3 = 18 ...(2)
Also we know
an = a + (n - 1)d
For the 2nd term (n = 2)
a2 = a + (2 – 1)d
14 = a + d (Using 1)
a = 14 – d ...(3)
Similarly for the 3rd term (n = 3)
a3 = a + (3 - 1)d
18 = a + 2d ...(Using 2)
a = 18 – 2d ...(4)
Subtracting (3) from (4), we get,
a – a = (18 – 2d) – (14 – d)
0 = 18 – 2d – 14 + d
0 = 4 – d
d = 4
Now, to find a, we substitute the value of d in (4),
a = 14 – 4
a = 10
So for the given A.P d = 4 and a = 10
So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,
Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula for n = 51, we get,
S51 = `51/2 [2(10) + (51 - 1)(4)]`
= `51/2 [20 + (50)(4)]`
= `51/2 [20 + 200]`
= `51/2 [220]`
= 5610
Therefore, the sum of first 51 terms for the given A.P is S51 = 5610.
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