Question

1 If the nth term of an A.P. is
7-3n. Find the sum of 30
terms​

Answer 1

Answer:

- 13740

Step-by-step explanation:

1st term = 7 - 30(1) = - 23

30th term = 7 - 30(30) = - 893

Using S = (n/2)(a + l), where a is first term and l is last term.

=> S = (30/2)(-23 + (-893))

= 15(-23 - 893)

= 15(- 916)

= - 13740

Answer 2

Answer:-

$\pink{\bigstar}$ Sum of 30 terms of the A.P $\large\leadsto\boxed{\rm\purple{-1185}}$

• Given:-

nth term of an A.P = 7-3n

• To Find:-

Sum of 30 terms

• Solution:-

Given that,

$\sf{a_n = 7 - 3n}$

$\pink{\bigstar}$ First term [a]:-

→ $\sf{a_1 = 7 - 3(1)}$

→ $\sf{a_1 = 7 - 3}$

→ $\bf{a_1 = 4}$

$\pink{\bigstar}$ Second term:-

→ $\sf{a_2 = 7 - 3(2)}$

→ $\sf{a_2 = 7 - 6)}$

→ $\bf{a_2 = 1}$

$\pink{\bigstar}$ Difference [d]:-

$\sf{d = a_2 - a_1}$

→ $\sf{d = 1 - 4}$

→ $\bf{d = -3}$

$\pink{\bigstar}$ 30th term [l]:-

$\large\boxed{\bf\green{a_{n} = a + (n - 1)d}}$

→ $\sf{a_{30} = 4 + (30 -1) \times -3}$

→ $\sf{a_{30} = 4 + 29 \times - 3}$

→ $\sf{a_{30} = 4 + (-87)}$

→ $\sf{a_{30} = 4 - 87}$

→ $\bf{a_{30} = -83}$

$\red{\bigstar}$ Sum of 30 terms:-

$\large\boxed{\bf\green{S_n = \dfrac{n}{2} [a + l]}}$

→ $\sf{S_{30} = \dfrac{30}{2} [4 + (-83)]}$

→ $\sf{S_{30} = 15 [4 - 83]}$

→ $\sf{S_{30} = 15 \times -79}$

→ $\bf\green{S_{30} = -1185}$

Therefore, the sum of the A.P is -1185.