Question

If the sum of the first 14 terms of an A.P. is 1050 and its first term is 10, then it 19th
term is how much​

Answer 1

Given :-

If  the sum of the 14 terms of an A.P. is 1050 and its first term is 10

To Find :-

19 th term

Solution :-

S₁₄ = 1050

We know that

Sₙ = n/2[2a + (n - 1)d]

1050 = 14/2[2(10) + (14 - 1)d]

1050 = 7[20 + 13d]

1050/7 = 20 + 13d

150 = 20 + 13d

150 - 20 = 13d

130 = 13d

130/13 = d

10 = d

Now

aₙ = a + (n - 1)d

a₁₉ = 10 + (19 - 1)10

a₁₉ = 10 + (18)10

a₁₉ = 10 + 180

a₁₉ = 190

[tex][/tex]

Answer 2

Answer:

here 1st term a=10

common difference=d

sum of 14 terms=1050=14/2{2a+(n-1).d}

=> 1050=7{2.10+(14-1).d}

=> 1050=7(20+13d)

=> 1050/7=20+13d

=> 150=20+13d

=> 13d=150-20

=> 13d=130

=> d=130/13

=> d=10

so , common difference=10

so 19 th term

=a+(19-1).d

=10+18.10

=10+180

=190