The nth term of an A.P. is given by (-4n + 15). Find the sum of first

1.	The nth term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P.
Answer» Given: The n^thterm of an A.P. (-4n+15). To find: the sum of first 20 terms of this A.P. Solution: We have, T_n = (-4n + 15) T₁ = -4 + 15 = 11 T₂ =( -4 × 2 )+ 15 = 7 T₃ =( -4 × 3 ) + 15 = 3 Hence the A.P is 11,7,3,......... The first term is 11 and the common difference is, d = T₂ – T₁ = 7 – 11 = -4 We calculate the sum of terms of A.P by using the formula:S_n= \(\frac{n}{2}\)[2a + (n – 1) d] Substitute the known values to get the sum. The sum of first 20 terms, S₂₀ = \(\frac{20}{2}\)[(2 x 11) + (20 - 1)(-4)] S₂₀ = \(\frac{20}{2}[(22 + (19)(-4)]\) S₂₀ = 10 (22 – 76) S₂₀ = 10(-54) S₂₀ = -540

The nth term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P.

Answer»

Given:

The n^thterm of an A.P. (-4n+15).

To find: the sum of first 20 terms of this A.P.

Solution: We have,

T_n = (-4n + 15)

T₁ = -4 + 15 = 11

T₂ =( -4 × 2 )+ 15 = 7

T₃ =( -4 × 3 ) + 15 = 3

Hence the A.P is 11,7,3,.........

The first term is 11 and the common difference is, d = T₂ – T₁ = 7 – 11 = -4

We calculate the sum of terms of A.P by using the formula:S_n= \(\frac{n}{2}\)[2a + (n – 1) d]

Substitute the known values to get the sum.

The sum of first 20 terms, S₂₀ = \(\frac{20}{2}\)[(2 x 11) + (20 - 1)(-4)]

S₂₀ = \(\frac{20}{2}[(22 + (19)(-4)]\)

S₂₀ = 10 (22 – 76)

S₂₀ = 10(-54)

S₂₀ = -540