Question

If the case of the sum of the first 4terms=the sum of the 13term=78

Answer 1

Answer:

n = 13, Sum(13) = 21

Sum = (n/2)( 2a + (n-1)d)

21 = (13/2)( 2a + (13-1)d))

42/13 = (2a + (12)d)

n = 21, Sum (21) = 13

Sum = (n/2)( 2a + (n-1)d)

13 = (21/2)( 2a + (21-1)d)

26/21 = (2a + (20)d) → 13/21 = a + 10d - equation 1

42/13 = (2a + (12)d) → 21/13 = a + 6d - equation 2

equation 1 - equation 2 gives

4d = (13/21) - (21/13)

4d = (169–441)/273

4d = -272/273

d = -68/273

Substituting value of d in equation 1 →

a + 10(-68/273) = 13/21

a = (13/21) + (680)/273

a = ((169+ 680)/273)

a = 849/273

a = 283/91 ( dividing numerator and denominator by 3)

Now, n= 34

Sum = (n/2)( 2a + (n-1)d)

Sum(34) = (34/2)(2(283/91) + (34-1)(-68/273))

Sum(34) = (17)((566/91) - (33*68)/273))

Sum(34) = (17)((566/91) - (2244)/273))

Sum(34) = (17)((566*3/91*3) + (2244)/273))

Sum(34) = (17)((1698/273) + (2244)/273))

Sum(34) = (17)((1698 - 2244)/273))

Sum(34) = (17)((-546)/273))

Sum(34) = (17)(-2)

Sum(34) = -34

Therefore, the sum of the first 34 terms of this series is -34.

Hope this helps.

Step-by-step explanation: