Question

How to find the 1st term, the common difference, and the nth term of the arithmetic sequence described below? (1) 4th term is 11; 10th term is 29 (2) 8th term is 4; 18th term is -96

Answer 1

1)`1` st term is `2` and common difference is `3`
2)`1`st term is `74` and common difference is `-10`
`n` th term of an A.P series is `T_n=a+(n-1)d`

Explanation:

Let `a,d,n` be the first term , common difference and number of

terms respectively on A.P series.

`n` th term of an A.P series is `T_n=a+(n-1)d`

1) `T_4=11 :. a+(4-1)d = 11 or a +3 d = 11; (1)`

`T_10=29 :. a+(10-1)d = 29 or a +9 d = 29; (2)`

Multiplying equation (1) by `3` on both sides we get,

`3 a +9 d = 33; (3)` . Subtracting equation (2) from equation (3)

we get , `(3 a +9 d)- (a+9 d) = 29 or 2 a =4 or a =2` .

Putting `a=2` in equation (1) we get, `3 d= 11-2 or d =3`

`1` st term is `2` and common difference is `3`

2) `T_8=4 :. a+(8-1)d = 4 or a +7 d = 4; (1)`

`T_18=-96 :. a+(18-1)d = -96 or a +17 d = -96; (2)`

Subtracting equation (2) from equation (1) we get,

`-10 d =100 or d = -10` Putting `d=-10` in equation (1)

we get, `a +7 *(-10)=4 or a =74`

`1` st term is `74` and common difference is `-10` [Ans]

Answer 2

`a_4=11`
`a_10=29`

There are many ways to resolve. One of them is apply definition of general term of an arithemtic sequence

`a_n=a_1+(n-1)d`

We have
`a_4=a_1+(4-1)d=a_1+3d=11`
`a_10=a_1+(10-1)d=a_1+9d=29`

With this pair of values we have enough to determine `a_1 and d`

`a_1=11-3d`, then in second, `11-3d+9d=29` or equivalent

`6d=29-11=18` so `d=18/6=3` and `a_1=11-3·3=2`

We have then `a_n=2+(n-1)3=3n-1=`

Lets check if this general term works

`a_10=2+9·3=29`

Now follow the same process to resolve second sequence
`a_8=4`
`a_18=-96`