Question

Find the sum of first 21 terms of the ap whose 2nd term is 8 and 4 term is 14.

Answer 1

In AS( arithmetic sequence ),

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, $a_{l}= 21 th term$

We know that the sum of n terms from the first term is $\dfrac{n}{2}(a + a_{l}$ where $a_{l})$ is the last term of the AP.

Now, substituting the given & solved values.

⇒ Sum of 21 terms = $\dfrac{21}{2}( 5 + 65 )$

⇒ Sum of 21 terms = $\dfrac{21}{2}(70)$

⇒ Sum of 21 terms = 21( 35 )

⇒ Sum of 21 terms =  735

Therefore the sum of first 21 terms of the AP is 735.

Answer 2

Hey frnd Here is your answer:

It is given that , a2= 8 and a4= 14
Now , We know that an= a + (n-1)d
So Applying this formula we get,
a2= a + (2-1)d
a + d = 8 ... .. . . . . . .(1)
a4= a + (4-1)d
a + (4-1)d = 14
a + 3d = 14 .. . . .. ... .. (2)
Soving these eqn we will get a , d .
So ,We now we Have : a =5 d= 3 n= 21
Now, Sn = n/2 (2a + (n-1)d )
S21= 21 / 2 (2 * 5 + (21-1)3
= 21/ 2 (10+ 20*3)
= 21 / 2 (10+60)
= 21 / 2 (70) 35
S21= 21 * 35
S21= 735
Here is your answer: = 735
Hope this helps☺☺