Question

for an A.P.T6=47 and T10=10 then find T30.

Answer 1

so let a be first term and d be common difference 

we have t6 = a + 5d = 47 ..... i

and t10 = a + 9d = 10 .......... ii

subtracting i from ii 

- 4d = 37 

d = - 37/4

so a = 93.25 

so t30 = a + 29d = - 175 ANSWER

Answer 2

The general form of AP = a, a+d, a+2d.....

Tn=a+(n-1)d

T6 = a+5d = 47.......(1)

T10 = a+9d = 10........(2)

Subtracting both,

-4d = 37

d = -$\frac{37}{4}$

Substituting d value in (1)

a + 5 $\frac{-37}{4}$ = 47

a = $47 + \frac{185}{4}$

a=93.25

T30 = 93.25 + (30-1) $\frac{-37}{4}$

T30 = 93.25 - 268.25

T30 = - 175