Question

the sum of the first n terms of an A.P is 2n. If the sum of the first 2n terms of this A.P is 3n, what will be the sum of the first 3n terms of the A.P?

Answer 1

Answer:

3n

Step-by-step explanation:

Let the first term be a and common difference be d.

In APs, sum = (n/2) {2a + (n - 1)d}

For first n terms:

= > 2n = (n/2) { 2a + (n - 1)d }

= > 2 * 2 = 2a + (n - 1)d

= > 4 = 2a + (n - 1)d

For first 2n terms:

= > 3n = (2n/2) { 2a + (2n - 1)d }

= > 3 = 2a + (2n - 1)d

= > 3 = 2a + (n - 1)d + nd

= > 3 = 4 + nd

= > - 1 = nd

For first 3n terms:

= > S = (3n/2) { 2a + (3n - 1)d }

= > S = (3n/2) { 2a + (n - 1)d + 2nd }

= > S = (3n/2) { 4 + 2(-1) }

= > S = (3n/2) { 4 - 2 }

= > S = 3n/2 * 2

= > S = 3n

Answer 2

Given:-

Sₙ = 2n = n/2{2a+(n-1) d}

S₂ₙ = 3n = 2n/2{2a+(2n-1) d}

S₂ₙ = 3n = n{2a+(2n-1) d}

Solution:-

✰ Solving Sₙ :-

Sₙ = 2n = n/2{2a+(n-1) d}

2n = n/2{2a+(n-1) d}

4 = 2a+(n-1) d _________(1)

✰ Solving S₂ₙ :-

S₂ₙ = 3n = n{2a+(2n-1) d}

3n = n{2a+(2n-1) d}

3 = 2a+(2n-1) d

3 = 2a+2nd-d

3 = 2a+nd+nd-d

3 = 2a + (n-1) d+nd

3 = 4 + nd ___________ (From 1)

-1 = nd ______________ (2)

✰ Solving S₃ₙ :-

S₃ₙ = 3n/2{2a+(3n-1)d}

S₃ₙ = 3n/2{2a+3nd-d}

S₃ₙ = 3n/2{2a+nd+2nd+nd-d}

S₃ₙ = 3n/2{2a+(n-1)d+2nd}

S₃ₙ = 3n/2{4+2×(-1)} _____(From 1&2)

S₃ₙ = 3n/2{4-2}

S₃ₙ = 3n/2{2}

S₃ₙ = 3n/2×2

S₃ₙ = 3n

S₃ₙ = 3n

Sum of first 3n terms is 3n.