Question

find the sum of first 51 term of an arithmetic progression whose second term is 14 and third term is 18 respectively ​

Answer 1

Answer:

Solution:

a2 =14 and a3=18

Common difference, d = a3 − a2 = 18 − 14 = 4

Now

a2=a+d=14

a+4=14

a=10

Now sum of 51 terms

$sn = \frac{n}{2}(2a + (n - 1)d)$

$s51 = \frac{51}{2} (2(10) + (51 - 1)4)$

$= \frac{51}{2} (20 + 200)$

=$\frac{51 \times 220}{2}$

$= 51 \times 110$

$s51 = 5610$

Answer 2

Answer:

Acha 5:00pe aegi na

please

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Step-by-step explanation:

Sum of the first n terms of an AP is given by Sₙ = n/2 [2a + (n - 1) d] or Sₙ = n/2 [a + l], and the nth term of an AP is aₙ = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

Given,

2nd term, a₂ = 14

3rd term, a₃= 18

Common difference, d = a₃ - a₂ = 18 - 14 = 4

We know that nth term of an AP is, aₙ = a + (n - 1)d

a₂ = a + d

14 = a + 4

a = 10

Sum of n terms of AP is given by Sₙ = n/2 [2a + (n - 1) d]

S₅₁ = 51/2 [2 × 10 + (51 - 1) 4]

= 51/2 [20 + 50 × 4]

= 51/2 × 220

= 51 × 110

= 5610

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