Question

If the sum of first 100 terms of an AP is -1 and the sum of the even terms lying in the first 100 term is 1. then find first term , common difference and 100th term of AP.

Answer 1

first term = -74/25 ,common difference = 3/50 and 100th term of AP = 149/50

Step-by-step explanation:

AP

a , a+d , a+2d, ...................., a+ 99d

sum.= (100/2)(a + a + 99d)

= 100a + 50×99d

100a + 4950d = -1

a+d , a +3d , ..............., a + 99d

sun = (50/2)(a + d + a + 99d)

= 50a + 2500d

50a + 2500d = 1

100a + 5000d = 2

50d = 3

d = 3/50

50a + 2500(3/50) = 2

50a + 150 = 2

50a = -148

25a = - 74

a = -74/25

a + 99d

= -74/25 + 99(3/50)

= (-148 + 297)/50

= 149/50

first term = -74/25

common difference = 3/50

and 100th term of AP = 149/50

learn more:

In an A.P if sum of its first n terms is 3n square +5n and it's Kth term ...

https://brainly.in/question/8236011

Find the sum of frist 51 terms of the AP whose 2nd term is 2 and 4th ...

https://brainly.in/question/7655866

Answer 2

Answer:

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) d

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56⇒ 4(11) + 6d = 56 [Since a = 11 (given)]

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56⇒ 4(11) + 6d = 56 [Since a = 11 (given)]⇒ 6d = 12

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56⇒ 4(11) + 6d = 56 [Since a = 11 (given)]⇒ 6d = 12⇒ d = 2

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56⇒ 4(11) + 6d = 56 [Since a = 11 (given)]⇒ 6d = 12⇒ d = 2∴ 4a + (4n –10) d = 112

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56⇒ 4(11) + 6d = 56 [Since a = 11 (given)]⇒ 6d = 12⇒ d = 2∴ 4a + (4n –10) d = 112⇒ 4(11) + (4n – 10)2 = 112

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56⇒ 4(11) + 6d = 56 [Since a = 11 (given)]⇒ 6d = 12⇒ d = 2∴ 4a + (4n –10) d = 112⇒ 4(11) + (4n – 10)2 = 112⇒ (4n – 10)2 = 68

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56⇒ 4(11) + 6d = 56 [Since a = 11 (given)]⇒ 6d = 12⇒ d = 2∴ 4a + (4n –10) d = 112⇒ 4(11) + (4n – 10)2 = 112⇒ (4n – 10)2 = 68⇒ 4n – 10 = 34

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56⇒ 4(11) + 6d = 56 [Since a = 11 (given)]⇒ 6d = 12⇒ d = 2∴ 4a + (4n –10) d = 112⇒ 4(11) + (4n – 10)2 = 112⇒ (4n – 10)2 = 68⇒ 4n – 10 = 34⇒ 4n = 44

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6dSum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]= 4a + (4n – 10) dAccording to the given condition,4a + 6d = 56⇒ 4(11) + 6d = 56 [Since a = 11 (given)]⇒ 6d = 12⇒ d = 2∴ 4a + (4n –10) d = 112⇒ 4(11) + (4n – 10)2 = 112⇒ (4n – 10)2 = 68⇒ 4n – 10 = 34⇒ 4n = 44⇒ n = 11