Question

the 6th term of an ap is 11 and the 21st term is 41. Find the sum of first 30 terms ​

Answer 1

ANSWER:

SUM OF 30 TERMS = 900

GIVEN:

6th term of AP = 11

21st term of AP= 41

TO FIND:

Sum of first 30 terms

SOLUTION:

nth term of an AP = a+(n-1)d

6th term= 11

$\implies \: a + (6 - 1)d = 11 \\ \implies \: a + 5d = 11 \: \: \: \: \: \: \: .....(i)$

21st term = 41

$\implies \: a + (21 -1)d = 41 \\ \implies \: a + 20d = 41 \: \: \: \: \: \: ........(ii)$

Subtracting eq (i) from (ii) we get;

$\implies \: 15d = 30 \\ \implies \: d = \frac{30}{15} \\ \implies \: d = 2$

putting d= 2 in eq (i) we get;

$\implies \: a + 5 \times 2 = 11 \\ \implies \: a = 11 - 10 \\ \implies \: a = 1$

Now sum of 30 terms:

Sum of n terms of an AP is given by:

$sum \: (n \: terms) = \frac{n}{2} (2a + (n - 1)d)$

Using this formula:

$sum = \frac{30}{2} (2 \times 1 + (30 - 1)2) \\ = 15(2 + 58) \\ = 15(60) \\ = 900$

SUM OF 30 TERMS = 900