Question

Find the no. Of term of the A.P. -12,-9,-6....,21 if 1 is added to each term of this A.P. ,then find the sum of all terms of the A.P. thus obtained.

Answer 1

So we have A.P.,

→ - 12, - 9, - 6,......, 21

• aₙ = a + (n - 1)d

→ 21 = - 12 + (n - 1)3

→ 33 = 3(n - 1)

→ 11 + 1 = n

→ 12 = n

Hence there are 12 terms in A.P.

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If 1 is add in each term then A.P will be,

→ - 11, - 8, - 5,......, 22

• Sₙ = n/2 [2a + (n - 1)d]

→ Sₙ = 12/2 × [2(-11) + (12 - 1)3]

→ Sₙ = 6[- 22 + 33]

→ Sₙ = 6[11]

→ Sₙ = 66

Hence sum of 12 terms is 66.

Answer → 12 and 66

Answer 2

AP = -12, - 9, - 6 ,......, 21

Using formula we get :-

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

21 = -12 + (n - 1) × 3

33 = 3(n - 1)

${\implies\dfrac{33}{3} = 3(n - 1)}$

11 = n - 1

11 + 1 = n

12 = n

Therefore we get :-

12 terms in A.P.

Now according to second condition

When 1 is added to each term of A.P

= - 11, - 8, - 5,......, 22

Sum of terms of AP

${\implies S_{n} = \dfrac{n}{2}[2a + (n - 1)d}$

${\implies S_{n} = \dfrac{12}{2}[2(-11) + (12 - 1)\times 3}$

${\implies S_{n} = 6 (-22 + 33)}$

${\implies S_{n} = 6\times 11}$

${\implies S_{n} = 66}$

Therefore :-

Sum of 12 terms is 66.