Question

1. The fourth terms of an AP is 10 and the eleventh term of it exceeds three times the fourth term by one. Find the sum of the first twenty terms of the AP.​

Answer 1

${\huge{\underline{\large{\mathbb{\green{VERIFIED \: ANSWER}}}}}}$

The 4th term is a + 3d = 10 … (1)

The 11th term is (a + 10d) - 3(a + 3d) = 1, or

a + 10d - 3a - 9d = 1, or

-2a + d = 1 … (2)

From (1) a = 10 - 3d. Put that in (2) to get

-2(10 - 3d) + d = 1, or

-20 + 6d + d = 1, or

7d = 21, or

d = 3.

From (2) d-1 = 2a, 3–1 = 2 = 2a, or a = 1

Sn = (n/2)[2a + (n-1)d], or

S20 = (20/2)[2*1 + (20–1)*3]

= 10[2 + 19*3]

= 10(2+57)

= 10*59

= 590

Answer = 590.

Answer 2

$\sf{a_4=10}$

$\sf{a_{11}=3a_4+1}$

To find :- $\sf{S_{20}}$

$\sf{\boxed{\bold{\footnotesize{a_n=a_1+(n-1)d }}}}$

From $\sf{a_4}$ and $\sf{a_{11}}$

a + 3d = 10 ______ ( 1 )

a + 10d = 3( a + 3d ) + 1

→ d = 2a + 1 ______ (2)

Putting (2) in ( 1 )

a + 3( 2a + 1 ) = 10

→ a = 1

So , d = 3

Now , $\sf{{S_n = {\dfrac{n}{2}} (2a+(n-1)d)}}$

$\sf{{S_{20}= {\dfrac{20}{2}} ( 2 \times(1) + (20-1)\times 3)}}$

→ $\sf {{S_{20}} = 590}$