Question

3. Find the sum of 28 terms of an A.P. whose nth
term is 8n - 5.​

Answer 1

Step-by-step explanation:

1sst term = (8×1)-5=3

2nd term = (8×2)-5=11

d=t2-t1

=11-3

=8

28th term = (8×28)-5=219

Sn = n/2[2a+(n-1)d]

= 14[6+(27×8)]

= 14(6+216)

= 14(222)

= 3108

Answer 2

Answer:

• Sum of 28 terms of AP is 3108.

Step-by-step explanation:

Given:

• nth term of AP = 8n - 5

To Find:

• Sum of 28 terms

Now, we know that,

⇒ aₙ = 8n - 5

⇒ a₁ = 8(1) - 5

⇒ a₁ = 8 - 5

⇒ a₁ = 3

⇒ aₙ = 8n - 5

⇒ a₂ = 8(2) - 5

⇒ a₂ = 16 - 5

⇒ a₂ = 11

Now, common difference, d = a₂ - a₁

⇒ d = 11 - 3

⇒ d = 8

Now, we will calculate sum of 28 terms,

$\implies \sf S_{n}=\dfrac{n}{2}[2a+(n-1)d]\\ \\ \\ \implies \sf S_{28}=\dfrac{28}{2}[6+(28-1)8]\\ \\ \\ \implies \sf S_{28}=14[6+(27)8]\\ \\ \\ \implies \sf S_{28}=14[6+216]\\ \\ \\ \implies \sf S_{28}=14[222]\\ \\ \\ \implies \sf S_{28}=3108$

Hence, Sum of 28 terms of AP is 3108.