Question

Find the sum of AP: (i)3, 7, 11, ... upto 35 terms. (ii)6+13+20+...+97​

Answer 1

Answer:

(i) 2481

(ii) 721

is the sum of A. p. for both the questions

I hope it is helpful to you.

Answer 2

i) Given ,

• First term (a) = 3
• Common difference (d) = 4
• Number of terms (n) = 35

We know that , the sum of first n terms of an AP is given by

$</p><p> \boxed{ \tt{S_{n} = \frac{n}{2} \{a + (n - 1)d \}}} \: \: \tt{or} \: \: </p><p> \boxed{ \tt{S_{n} = \frac{n}{2} \{a + a_{n}\}}}$

Thus ,

$\tt \implies S_{35} = \frac{35}{2} (2 \times 3 + (35 - 1)4 \}$

$\tt \implies S_{35} = \frac{35}{2} \{ 6 + 34 \times 4\}$

$\tt \implies S_{35} = \frac{35}{2} \times 142$

$\tt \implies S_{35} =35 \times 71$

$\tt \implies S_{35} =2485$

(ii) Given ,

• First term (a) = 6
• Common difference (d) = 7
• Nth term (an) = 97

We know that , the nth term of an AP is given by

$\boxed{ \tt{ a_{n} = a + (n - 1)d} }$

Thus ,

97 = 6 + (n - 1)7

91 = (n - 1)7

13 = n - 1

n = 14

$\therefore$ The sum of first 14th terms of AP will be

$\tt \implies S_{14} = \frac{14}{2} (6 + 97)$

$\tt \implies S_{14} =7 \times 103$

$\tt \implies S_{14} =721$