Question

34+32+30+--------+10

Answer 1

a = a₁ = 34 ; a₂ = 32 ; a₃ = 30

a₂-a₁ = 32-34 = -2
a₃-a₂ = 30-32 = -2
a₂-a₁ = a₃-a₂ =...

Therefore,the given series are in AP
Common difference,d = -2

Let 10 be nth term.
an = a + (n-1)d
10 = 34 + (n-1)(-2)
10-34 = (n-1)(-2)
-24 = (n-1)(-2)
n-1 = -24/-2
n-1 = 12
n = 13

Therefore, there are 13 terms in the given series.(AP)

Sn = $\frac{n}{2}$ [a + l ]
= $\frac{13}{2}$ (34+10)
= $\frac{13}{2}$ (44)
= 13(22)
= 286

The sum of the given series in AP = 286

Answer 2

Answer:-

$\small\sf{a=34}$ and $\small\sf{d=32-34=-2}$

Let the nth term of the AP is 10

$\large\sf{Therefore,}$

$\small\sf{an=10}$

$\small\sf{a+(n-1)d=10}$

$\small\sf{34+(n-1)(-2)=10}$

$\small\sf{(n-1)(-2)=-24}$

$\small\sf{n-1=12}$

$\small\sf{n=13}$

The sum of n terms of and AP is given by

$\small\sf{Sn=\frac{n}{2}(a+l)}$

$\small\sf{S13=\frac{13}{2}(34+10)}$

$\small\sf{S13=\frac{13}{2}(44)}$

$\small\sf{S13=13×22}$

$\small\sf{S13=286}$