Question

Ap has 40 term finnd the 40 term of ap 8 10 12. Iso find the sum of last 10 term

Answer 1

Given,

AP : 8, 10, 12,...

To Find out :

i) 40th term of the given AP series

ii) Sum of last 10 terms

Solution :

i) Given, a = 8

d = $a_{2} - a_{1}$

d = 10 - 8

d = 2

n = 40

Putting the given values in the formula $\boxed{\mathbf{a_{n} = a + (n - 1) d}}$

$a_{40} = 8 + (40 - 1)(2) \\ a_{40} = 8 + (39)(2) \\ a_{40} = 8 + 78 \\ \bf a_{40} = 86$

ii) Sum of $\bf last$ 10 terms of AP :

Here,

a = 86 [since we have to add up last 10 numbers]

d = - 2

n = 10

Putting the values in the formula $\boxed{\mathbf{S_{n} = \frac{n}{2}(2a + (n - 1)d)}}$

$S_{10} = \frac{10}{2}(2(86) + (10 - 1)(-2) \\ S_{10} = 5(172 + (9)( -2) ) \\ S_{10} = 5(172 - 18) \\ S_{10} = 5(154) \\ \bf S_{10} = 770$

Hope it helps! ;)

Answer 2

Answer:

Step-by-step explanation:Tn=40

d=10-8=2

T1=a=8

We know that ,

Tn=a+(n-1)d

T40=8+(40-1)*2

T40=8+39*2

T40=8+78

T40=86

'.' the 40th term is 86

The sum of last ten terms is

T31= 86-10d

86-10*2

86-20

66

'.' here a= 66= t1

Tn=86

We know that

Sn=n/2(t1+tn)

S10=10/2(66+86)

S10=5*152

S10=760

'.' sum of last 10 terms will be 760