Question

t19 = 52 , t38= 148, S56=?​

Answer 1

Step-by-step explanation:

t19=a+(19-1)d=52

=a+18d=52...............(1)

t38=a+(38-1)d=148

=a+37d=148...........(2)

solve equation 1and 2...

we get,

a+37d-a-18d=148-52

=19d=. 96

...d=96/19=5.05....

then solve a=?

then S56=56/2(2a+(n-1)d)...

put the value if a and d then solve it...

Answer 2

Answer:

Step-by-step explanation: T19=a+(n-1)d

52=a+18d....eq`n (¡)

T38=a+(n-1)d

148=a+37d.....eq`n(¡¡)

Solve it to find the value of d=5.05

Putting (d =5.05) in eq`n (i)

a=-43.90

S=n/2[2a+(n-1)d]

S=56/2(365.55)

S= 10,235.4