Question

Find the terms of an ap whose nth terms is (5-2n).Hence find the sum of its first 20 terms

Answer 1

since

Tn=5-2n,

then

T1=5-2×1=3,

T2=5-2×2=1,

T3=5-2×3=-1,

here
a=3,

d=T2-T1=1-3=-2,

n=20,

therefore

Sum of 20 terms=20/2 [2×3+(20-1)×(-2)],

=10[6-19×2],

=10[6-38],

=10×-32,

=-320

Answer 2

Answer:

The sum of first 20 terms of the given AP is -320

Step-by-step explanation:

• Arithmetic progression (AP) is a series in which every consecutive term has the same difference which we call as common difference denoted by 'd' and the term of series is generally denoted by $a_{n}$.
• Here we are provided with a formula 5-2n that will be used to generate the AP.
• For every natural number 'n' we get a term of AP
• Here we are asked to obtain the sum of first 20 terms that will be given by

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

       where n is number of terms

                  a is 1st term

                 d is common difference

Step 1:

• First we must obtain all the values which we require to obtain the sum namely n,a,d
• Since we need sum of first 20 terms so n will be 20
• Now to obtain a put n=1 (first term) in 5-2n, we get

        5-2(1) = 5-2 = 3

        a=3

• Now to obtain d we must find 2nd term of the series.

        Put n=2

        We get 5 - 2 (2) = 5 - 4 = 1

        So the common difference will be the difference of a term and its               preceding term

       We get 1 - 3 = -2

       d = -2

Step 2:

Substitute the values in       $S_{n} = \frac{n}{2}(2a+(n-1)d)$

We get

$S_{20} = \frac{20}{2}(2(3)+(20-1)(-2))$

$S_{20} = 10(6 - 19 * 2)$

$S_{20} = 10(-32)$

$S_{20} = -320$

Therefore the required sum of first 20 terms is -320