Question

Q17. The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120,
then find the number of terms and the common difference.
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Answer 1

Answer:

6   ; 10

Step-by-step explanation:

  Let there are x terms:

Let the first term be 'a' and common difference be 'd'.

 a₁ = - 5     ;  aₓ = 45

    ⇒ a₁ + aₓ = -5 + 45 = 40

Sum of all terms = 120

(number of term/2)(1st + last term)=120

 (x/2) (a₁ + aₓ) = 120

 (x/2)( 40 ) = 120

 x = (120*2 /40) = 6

Number of terms = 6

   Last term - first term = 45 - (-5)

   aₓ - a = 45 + 5

   a + (6 - 1)d - a = 50

   5d = 50

      d = 10

Answer 2

Step-by-step explanation:

Sn=120

a=5

an=45

d=?

n=?

Sn=n/2(a+an)

120=n/2(5+45)

120=n/2(50)

120=25n

120/25=n

n=4.8

an=a+(n-1)d

45=5+(4.8-1)d

40=4.7d

40/4.7=d

d=8.5