Question

if the sum of the first 14 terms of an AP is 1050 and its first term is 10,find the 20th term

Answer 1

a=10...given

S14=1050
⇒n/2[2a+(n-1)d]=1050
⇒14/2[20+13d]=1050
⇒7[20+13d]=1050
⇒20+13d=1050/7
⇒20+13d=150
⇒13d=150-20
⇒13d=130
⇒d=10

20th term=a+19d
                 =10+19×10
                 =10+190
                 =200...Answer

I hope its help you..

Answer 2

Solution :

In this Question We have provided with first term of an A.P. and the sum of the 14 terms of an AP is 1050 . It implies we have n ( term of an AP ) = 14.

Using Sum of an AP formula for finding the common difference of an AP

$S_n = \frac{n}{2} [ 2a + ( n - 1) d ]$

⇒  $1050 = \frac{14}{2} [ 2 * 10 + ( 14 - 1 ) d ]$

⇒  $1050 = 7[20 + 13d]$

⇒ $\frac{1050}{7} = 20 + 13d$

⇒  $150 = 20 + 13d$

⇒  150 - 20 = 13d

⇒ 130 = 13d

⇒ d = 10

Therefore , the common difference of an AP is equal to 10

Finding the 20th term of an AP

a + ( n - 1) d

⇒ a + ( 20 - 1) d

⇒ a + 19 d

⇒10 + 19 * 10

⇒ 10 + 190

⇒ 200

Therefore , the 20th term of an AP is equal to 200