Find the sum of first 21 terms of an A.P whose 2nd term is 8 and 4th term is 14
Answers
Answer:
First term = a
Common difference = d
Second term = a₂ = a + d
Third term = a₄ = a + 3d
xth term = aₓ = a + ( x - 1 )d
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In the given question, 2nd term is 8 and 4th term is 14.
∴ a₂ = a + d = 8
⇒ a + d = 8
⇒ a = 8 - d ...( i )
Given, a₄ = a + 3d = 14
∴ a + 3d = 14
⇒ a = 14 - 3d ...( ii )
As both the equation ( i ) & ( ii ) are equal to a,
⇒ 8 - d = 14 - 3d
⇒ 3d - d = 14 - 8
⇒ 2d = 6
⇒ d = 6 / 2
⇒ d = 3
Substituting the value of d in ( i )
a = 8 - d
a = 8 - 3
a = 5
∴ 21 th term = a + ( 21 - 1 )d
= 5 + ( 20 ) 3
= 5+ ( 60 )
= 65
Let 21 th term be the last of the AS.
As 21 th term is the last of the AS,
We know that the sum of n terms from the first term is where is the last term of the AP.
Now, substituting the given & solved values.
⇒ Sum of 21 terms =
⇒ Sum of 21 terms =
⇒ Sum of 21 terms = 21( 35 )
⇒ Sum of 21 terms = 735
Explanation:
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