Answer :

To Find: First four terms of given series.

(i) Given: n^{th} term of series is (5n + 2)

Put n=1, 2, 3, 4 in n^{th} term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term

a_{1} = (51 + 2) = 7

a_{2} = (52 + 2) = 12

a_{3} = (53 + 2) = 17

a_{4} = (54 + 2) = 22

First four terms of given series is 7, 12,17,22

__ALTER__: When you find or you have first term (a or a_{1}) and second term (a_{2}) then find the difference (a_{2} - a_{1})

Now add this difference in last term to get the next term

For example a_{1}= 7 and a_{2}= 12, so difference is 12 - 5 = 7

Now a_{3} = 12 + 5 = 17, a_{4} = 17 + 5 = 22

(This method is only for A.P)

__NOTE__: When you have nth term in the form of (an + b)

Then common difference of this series is equal to a.

This type of series is called A.P (Arithmetic Progression)

(Where a, b are constant, and n is number of terms)

(ii) Given: n^{th} term of series is

Put n=1, 2, 3, 4 in n^{th} term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term.

a_{1} = =

a_{2} = =

a_{3} = =

a_{4} = =

First four terms of given series are , , ,

(iii) Given: n^{th} term of series is (–1)^{n–1} × 2^{n + 1}

Put n=1, 2, 3, 4 in n^{th} term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term.

a_{1} = (–1)^{1–1} × 2^{1 + 1} = (–1)^{0} × 2^{2} = 14 = 4

a_{2} = (–1)^{2–1} × 2^{2 + 1} = (–1)^{1} × 2^{3} = (–1)8 = (–8)

a_{3} = (–1)^{3–1} × 2^{3 + 1} = (–1)^{2} × 2^{4} = 116 = 16

a_{4} = (–1)^{4–1} × 2^{4 + 1} = (–1)^{3} × 2^{5} = (–1)32 = (–32)

First four terms of given series are 4, –8 , 16 ,–32

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