Question

find the 32nd terms of an AP 1,3,5,7,,,,257. From its last term of an AP​

Answer 1

Answer:

The given progression 9, 15, 21, 27, ... .

Clearly, 15 − 9 = 21 − 15 = 27 − 21 = 6 (Constant)

Thus, each term differs from its preceding term by 6. So, the given progression is an AP.

First term = 9

Common difference = 6

Next term of the AP = 27 + 6 = 33

(ii) The given progression 11, 6, 1, −4, ... .

Clearly, 6 − 11 = 1 − 6 = −4 − 1 = −5 (Constant)

Thus, each term differs from its preceding term by −5. So, the given progression is an AP.

First term = 11

Common difference = −5

Next term of the AP = −4 + (−5) = −9

(iii) The given progression −1,

-5

6

,

-2

3

,

-1

2

, ...

Clearly,

-5

6

-(-1)=

-2

3

-(

-5

6

)=

-1

2

-

Answer 2

Answer:

Answer is attached

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