Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Answers
Answer:
The sum of first 51 terms 5610 .
Step-by-step explanation:
Given :
a2 = 14 and a3 = 18
Case 1 :
By using the formula ,an = a + (n - 1)d
a2 = 14
a + (2 - 1)d = 14
a + d = 14
a = 14 – d…………... (1)
Case 2:
a3 = 18
a + (3 - 1)d = 18
a + 2d = 18
(14 - d) + 2d = 18
[From eq 1]
14 + d = 18
d = 18 - 14
d = 4
On Putting the value of d = 4 in eq (1),
a = 14 – d
a = 14 – 4
a = 10
By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]
S51 = 52/2 [2(a) + (51 - 1)(d)]
S51 = 51/2 [2 × 10 + 50 ×4]
S51 = 51/2 [20 + 200]
S51 = 51/2 × 220
S51 = 51 × 110
S51 = 5610
Hence, the sum of first 51 terms 5610 .
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GIVEN :
Second term of an AP = 14
a + d = 14 ------(1)
Third term of an AP = 18
a + 2d = 14
We know that the common difference between the terms is same.
a2 = a + d
a2 = 14
Common difference = a3 - a2
= 18 - 14
= 4
Common Difference = d = 4
Substitute d in eq - 1
a + d = 14
a + 4 = 14
a = 14 - 4
a = 10
First term = 10
In an AP sum of the terms = n/2 ( 2a + (n-1)d
S51 = 51/2 ( 2(10) + (51 - 1)4 )
S51 = 51/2 ( 20 + 200)
S51 = 51/2(220)
S51 = 51 × 110
Therefore, the sum of first 51 terms is 5610.