Question

In an ap 21 terms the sum of first 3term is -33 and that of middle 3 is 75 what is the sum of the ap ?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Sum\:of\:A.P=2037}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Number \: of \: terms = 21 \\ \\ \tt: \implies Sum \: of \: first \: 3 \: terms = - 33 \\ \\ \tt: \implies Sum \: of \: next \: three = 75 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Sum \:of \: A.P = ?$

• According to given question :

$\bold{For \: sum \: of \: first \: three \: term} \\ \tt: \implies s_{3} = \frac{3}{2}(2a + (3 - 1) \times d) \\ \\ \tt: \implies - 33 = \frac{3}{2} (2a + 2d) \\ \\ \tt: \implies - 33 = 3(a + d) \\ \\ \tt: \implies a + d = - 11 - - - - - (1) \\ \\ \bold{For \: next \: three \: sum \: of \: terms} \\ \tt: \implies 75 = \frac{3}{2} (2(a + 3d) + (3 - 1)d) \\ \\ \tt: \implies \frac{150}{3} = 2a + 6d + 2d \\ \\ \tt: \implies 50 = 2a + 8d \\ \\ \tt: \implies a + 4d = 25 - - - - - (2) \\ \\ \text{Subtracting \: (1) \: from \: (2)} \\ \tt: \implies 4d - d = 25 - ( - 11)$

$\tt: \implies 3d = 36 \\ \\ \tt: \implies d = \frac{36}{3} \\ \\ \green{\tt: \implies d = 12} \\ \\ \text{Putting \: value \: of \: d \: in \: (2)} \\ \tt: \implies a + 4 \times 12 = 25 \\ \\ \tt: \implies a + 48 = 25 \\ \\ \tt: \implies a = 25 - 48 \\ \\ \green{\tt: \implies a = - 23} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \\ \tt: \implies s_{21} = \frac{21}{2} (2 \times - 23 + (21 - 1) \times 12) \\ \\ \tt: \implies s_{21} = \frac{21}{2} ( - 46 + 240) \\ \\ \tt: \implies s_{21} = \frac{21}{2} \times 194 \\ \\ \green{\tt: \implies s_{21} =2037 }$

Answer 2

Answer:

Given:

• In an ap 21 terms the sum of first 3term is -33 and that of middle 3 is 75.

Find:

• What is the sum of the ap ?

According to the question:

• 21 = Total number of terms.

• -33 = Sum of first three terms.

• 75 = Sum of next three terms.

• Let us assume 'a₃' as the sum of first three sums and 'a₂₁' as total number of sums.

First three sums:

⇒ a₃ = 3/2 [2 (a + 3d) + (3 - 1) d]

⇒ -33 = 3/2 (2a + 2d)

⇒ -33 = 3 (a + d)

⇒ (a + d = 11)

⇒ -11 – Equation (1)

For next three terms:

⇒ 75 = 3/2 [2 (a + 3d) + (3 - 1) d]

⇒ 150/3 = (2a + 6d + 2d)

⇒ 50 = (2a + 8d)

⇒ (a + 4d)

⇒ 25 – Equation (2)

Subtracting equation (1) for equation (2):

⇒ 4d - d = 25 - (-11)

⇒ 3d = 36

⇒ d = 36/3 = 12

.•. d = 12

Adding values for 'd' in equation (2):

⇒ a + 4 × 12 = 25

⇒ a + 48 = 25

⇒ a = 25 - 48 = -23

.•. a = -23

Finding the sum of AP.

⇒ aₙ = n/2 [2a + (n - 1) d]

⇒ a₂₁ = 21/2 [2 × -23 + (21 - 1) × 12]

⇒ a₂₁ = 21/2 [-46 + 240]

⇒ a₂₁ = 21/2 × 194

⇒ a₂₁ = 2037

Therefore, 2037 is the sum of AP.