Question

the sum of the first 5 terms of an ap is 25 and sum of first 18 terms of an ap is 324, prove that the sum of its first n terms is n²​

Answer 1

Step-by-step explanation:

in the first case the number of terms is 5 and its square is 25 that is 5 square is 25 and in second case the number of terms are 18 and square of 18 is 324 so by this we can conclude that the sum of its first n terms is n square

and also we can say that the number of terms is 5

5 square equals to 25

and in second case we can say that the number of terms is 18

18 square = 324

by this we can conclude that the sum of its first n term is n square

Answer 2

Answer:

    Sₙ = n²

Step-by-step explanation:

Sₙ = n / 2 [2a + d(n-1)]

S₅ = 5 / 2 [2a + d(5-1)]

S₅ = 5 / 2 [2a + 4d]

25 = 5 / 2 [2a + 4d]

50  = 10a + 20d

5 = a + 2d -----{i}

Sₙ = n / 2 [2a + d(n-1)]

S₁₈ = 18 / 2 [2a + d(18-1)]

324 = 9(2a + 18d -d)

36 = 2a + 17d -----{ii}

5 = a + 2d -----{i} × -2

-10 = -2a + -4d -----{iii}

{ii} + {iii}

36 - 10 = 2a + 17d -2a + -4d

26 = 13d

d = 26 / 13

d = 2

5 = a + 2d

5 = a + 4

a = 1 , d = 2

Sₙ = n / 2 [2a + d(n-1)]

Sₙ = n / 2 [(2×1) + 2(n-1)]

Sₙ = n / 2 [2 + 2n - 2]

Sₙ = n / 2 [ 2n ]

Sₙ = n × n

Sₙ = n²

Hence proved