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Question

If the nth term of a progression is (4n-10) show that it is an AP. Find its (i) first term (ii) common difference, and (iii) 16th term.

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Solution

Given nth term of a progression is (4n10)

i.e., Tn=4n10

First term, T1=4×110=410=6=a

Second term, T2=4×210=810=2

Third term, T3=4×310=1210=2

T2T1=26=2+6=4

T3T2=22=2+2=4

i.e.,T2T1=T3T2=4=d, common difference

and therefore the sequence -6,-2,2,6,... is an AP.

16th term, T16=4×1610=6410=54


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