Question

the fourth term of an ap is 10 and 11th term of it exceeds three times the fourth term by 1 find the sum of the first 20 terms of arithmetic progression​

Answer 1

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.

$\huge\red{brainliest\:plz}$ ❤️

Answer 2

Answer:

Sum of first 20 term is 590.

Step-by-step explanation:

Given: 4th term of AP = 10

           11th term exceed 3 times 4th term by 1

To find: Sum of first 20 terms

Sum of n terms of AP is given by,

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

nth term of AP is given by,

$a_n=a+(n-1)d$

⇒ $a_4=a+(4-1)d=a+3d$

⇒ $a_{11}=a+(11-1)d=a+10d$

According to the question,

$a_4=10$

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

$a_{11}-3\times a_4=1$

a + 10d - 3 × 10 = 1

a + 10d - 30 = 1

a + 10d = 1 + 30

a + 10d = 31 ........ (2)

subtract (1)  from (2)

a + 10d - a - 3d = 31 - 10

7d = 21

d = 3 (by dividing 21 by 3)

put in eqn (1), we get

a + 3 × 3 = 10

a = 10 - 9

a = 1

$\implies S_{20}=\frac{20}{2}\times(2\times1+(20-1)\times3)$

$S_{20}=10\times(2+57)$

$S_{20}=10\times59$

$S_{20}=590$

Therefore, Sum of first 20 term is 590.