Question

8. Find the sum of first 15 terms of an A.P. whose nth term is 3 - 2n.
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Answer 1

Sn=3-2n

s1=3-2(1)

s1=3-2

s1=1

a=1

s2=3-2(2)

s2=3-4

s2=-1

tn=sn-sn-1

t2=1-1

t2=0

d=tn-tn-1

d=0-1

d=-1

sn=n/2[2a+(n-1)d]

s15=15/2[2(1)+15-1 -1]

s15=15/2[2+14[-1]]

s15=15/2[2-14]

s15=15/2[-12]

s15=15[-6]

s15=-90

HOPE THIS WILL HELP YOU

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Answer 2

Given:-

• nth term of the A.P. is 3-2n

To find:-

• Sum of first 15 term of the A.P = ?

Solution:-

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

It is given that,

$</u><u>T</u><u>_{n} = 3 - 2n$

So,

$= > T_{1} = 3 - 2(1) = 3 - 2 = 1$

$= > T_{2} = 3 - 2(2) = 3 - 4 = - 1$

Now,

a = 1

d = (-1-1)= -2

n= 15

We know that,

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

so that,

$= > S_{15} = \frac{15}{2} (2(1) + (15 - 1)( - 2) \\ = > S_{15} = \frac{15}{2} (2 + 14( - 2)) \\ = > S_{15} = \frac{15}{2} (2 - 28) \\ = > S_{15} = \frac{15}{2} \times ( - 26) \\ = > S_{n} = 15 \times ( - 13) = - 195$

Hence the sum of 15th term is -195

Hope its help uh❤