Question

in an auditorium there are 20 seats in the first row 26 seats in second row 32 in third row and 134 seat in last row find the total number of seats

Answer 1

The total number of seats in auditorium are 1540

Step-by-step explanation:

• Given

        Number of seat in different row are

        20 , 26, 32

        $\mathbf{First \ term(a_{1})=20}$

        $\mathbf{Second \ term(a_{2})=26}$

        $\mathbf{Third \ term(a_{3})=32}$

       $\mathbf{last \ term(l)=134=n^{th} \ term}$

       Total number of term(row) = n = unknown

• Here we see that given series are in A.P

       Common difference (d) =26-20 = 6

• From formula

       $\mathbf{n^{th}\ term\ (a_{n})=a_{1}+(n-1)d}$

       $\mathbf{134=20+(n-1)6}$

       $\mathbf{114=(n-1)6}$

       So

       Total number of row = n =20

• From formula of sum of n term

        $\mathbf{S_{n}=\frac{n}{2}(first \ term+ last \ term)}$

        $\mathbf{S_{n}=\frac{20}{2}(20+ 134)}$

       $\mathbf{S_{n}=10\times 154=1540}$ = This is total number of seat in auditorium.

Answer 2

Answer:

the answer is 818 seats

Step-by-step explanation:

Given

Tn=134

a=20

d=26-20

 =6

formula

Tn=a+(n-1)d

T134=20+(134-1)6

=20+(133)6

=20+798

=818

hence proved

pls mark brainliest