Question

The tenth 10th of term of an a.p is _37 nd the sum of the first six terms is _27 find the sum of its first eight terms​

Answer 1

from given

t10=37

a+9d=37

s6=27

6a+15d=27

2a+5d=9

next step is

Answer 2

Answer:

the correct answer is 4.

Step-by-step explanation:

beginning with "10th of term of an a.p is _37"

a10 = a + (10-1)d

37  = a + 9d .. (i)

now taking the second equation from " sum of the first six terms is -27 "

using the formula,

sum of n numbers = n/2 (a + a+ (n-1)d)

s6 = 6/2 (a+a+(6-1)d)

-27  = 3(2a + 5d)

(dividing the whole equation by 3 to simplify it)

-9 = (2a + 5d) ...(ii)

from (i) and (ii)

37  = a + 9d

-9 = (2a + 5d)

(multiplying i by 2)

74 = 2a + 9d

-9 = 2a + 5d

-----------------------

65 = 13d    (using elimination method)

d = 5

substituting the value of d in (ii)

-9 = 2a + 5(5)

-9 = 2a + (25)

-34 = 2a

a = -17

now, to find the sum of the first 8 terms

s8 = 8/2 (a + a+ (8-1) d)

    = 4 (2a + 7d)

   = 4 ( -34 + 35)

   = 4

thus the answer is 4.

TO CHECK:

AP : -17, -12, -7, -2, 3, 8  (for six terms)

if added = -27 (as given in the question.

AP : -17, -12, -7, -2, 3, 8, 13, 18 (for 8 terms)

if added = 4 (as solved above)

Thus checked.

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hope it helps:)