Question

An AP has 100 terms 1 term is 3 and 25 term is 75 then find out the sum of last 50 terms

Answer 1

Answer:

the sum of last 50 terms = -3525

Step-by-step explanation:

given that,

An AP has 100 terms and

1 term is 3 and 25 term is 75

let the common difference be d

given

first term(a) = 3

a25 = 75

so,

a + (25 - 1)d = 75

3 + 24d = 75

24d = 75 - 3

24d = 72

d = 72/24

d = 3

now,

to find,

the sum of last 50 terms

so,

here,

last term = its first term

last term

a + (50 - 1)d

3 + 49(3)

3 + 147

= 150

now,

here,

first term = 150

common difference = -3

so,

Sn = 50/2(2(3) + (50 - 1)(-3))

Sn = 25(6 + 49(-3))

= 25(6 - 147)

= 25(141)

= - 3525

so,

the sum of last 50 terms

= -3525

Answer 2

GIVEN :

An AP has 100 terms.

First term = a = 3

25th term = a + 24d = 75 -----(1)

Substitute a in eq - (1) to find common difference (d).

(3) + 24d = 75

24d = 75 - 3

24d = 72

d = 72/24

d = 3

First find the sum of 100 terms.

We know that,

Sum of terms in an AP is Sn = n/2 [ 2a + (n - 1)d ]

S100 = 100/2 [ 2(3) + (99)3]

S100 = 50 [ 6 + 297 ]

S100 = 50 [ 303 ]

S100 = 15150

Now find the sum of first 50 terms.

S50 = 50/2 [ 2(3) + (49)3 ]

= 25 [ 6 + 147 ]

= 25 [ 153 ]

= 3825

Now subtract the sum of first fifty terms from sum of 100 terms.

= 15150 - 3825

= 11325.

Therefore, Sum of last 50 terms is 11325.