Question

Find the 15th term of the A.P. If the 31st term is 40 and the sum of the 3rd
and 11th term is 16.

Answer 1

Answer :

15th term = 56/3

Step-by-step explanation :

         $\underline{\underline{\bf Arithmetic \ Progression :}}$

• It is the sequence of numbers such that the difference between any two successive numbers is constant.

• In AP,

         a - first term

         d - common difference

         aₙ - nth term

         Sₙ - sum of n terms

• General form of AP,

            a , a+d , a+2d , a+3d , ..........

• Formulae :-

             nth term of AP,

             $\boxed{\bf a_n=a+(n-1)d}$

__________________________

Given,

31st term = 40

sum of 3rd and 11th term = 16

• 31st term = 40

         $a+(31-1)d=40\\\\a+30d=40\\\\a=40-30d$

• Sum of 3rd term and 11th term = 16

           $a_3+a_{11}=16$

           $a+(3-1)d+a+(11-1)d=16\\a+2d+a+10d=16\\2a+12d=16\\2(a+6d)=2(8)\\a+6d=8\\\\a=8-6d$

40 - 30d = 8 - 6d

40 - 8 = 30d - 6d

32 = 24d

d = 32/24

d = 4/3

a = 8 - 6d

  = 8 - 6(4/3)

  = 8 - 2(4)

  = 8 - 8

  = 0

$\bigstar 15th \ term = 0+(15-1)(\frac{4}{3}) \\\\ 15th \ term=14(\frac{4}{3}) \\\\ 15th \ term=\frac{56}{3}$