Question

Find the sum of first 21 terms of an AP whose second term is 8 and forth term is 14​

Answer 1

Step-by-step explanation:

Given: a2 = 8 and a4 = 14 and n = 21

We know that,

a2 = a + d = 8 …(i)

and a4 = a + 3d = 14 …(ii)

Solving the linear equations (i) and (ii), we get

a + d – a – 3d = 8 – 14 ⇒ -2d = -6 ⇒ d = 3 Putting the value of d in eq. (i), we get

a + 3 = 8 ⇒ a = 8 – 3 = 5

Now,

we have to find the sum of first 21 terms.

⇒ S21 = 21 [5 + 10 × (3)]

⇒ S21 = 21 [35]

⇒ S21 = 735

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Answer 2

ANSWER:

Given:

• 2nd term of AP = 8
• 4th term of AP = 14

To Find:

• Sum of first 21 terms of the AP.

Solution:

$\text{We are given that,}\\\\:\implies a_2=8\:\:\:\:\&\:\:\:\:a_4=14\\\\\text{We know that,}\\\\:\implies a_n=a+(n-1)d\\\\\text{So,}\\\\:\implies a_2=a+(2-1)d=8\\\\:\implies 8=a+d- - - -(1)\\\\\text{Now,}\\\\:\implies a_4=a+(4-1)d=14\\\\:\implies14=a+3d- - - -(2)\\\\\text{To eliminate a, we subtract (1) from (2)}\\\\:\implies (14-8)=(a+3d)-(a+d)\\\\:\implies6=a\!\!\!/+3d-a\!\!\!/-d\\\\:\implies6=2d\\\\:\implies\underline{d=3}- - - -(3)\\\\\text{Putting value of d in (1). So,}\\\\:\implies8=a+(3)$

$:\implies a=8-3\\\\:\implies\underline{a=5}\\\\\text{We know that sum of n terms in an AP is,}\\\\:\longrightarrow S_n=\dfrac{n}{2}[2a+(n-1)d]\\\\\text{Here, n = 21, a = 5 and b = 3. So,}\\\\:\implies S_{21}=\dfrac{21}{2}\bigg(2(5)+(21-1)3\bigg)\\\\:\implies S_{21}=\dfrac{21}{2}\bigg(10+20(3)\bigg)\\\\:\implies S_{21}=\dfrac{21}{2}\bigg(10+60\bigg)\\\\:\implies S_{21}=\dfrac{21}{2\!\!\!/}(70\!\!\!\!/^{\:\:35})\\\\:\implies S_{21}=21*35\\\\\bf{:\implies S_{21}=735}$

$\text{\bf{\underline{Hence, sum of first 21 terms of the AP is 735.}}}$

Formulae Used:

• aₙ = a+(n-1)d
• Sₙ =(n/2)*(2a+(n-1)d)