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 $(4 n - 10)$

i.e., $T_{n} = 4 n - 10$

First term, $T_{1} = 4 \times 1 - 10 = 4 - 10 = - 6 = a$

Second term, $T_{2} = 4 \times 2 - 10 = 8 - 10 = - 2$

Third term, $T_{3} = 4 \times 3 - 10 = 12 - 10 = 2$

$T_{2} - T_{1} = - 2 - - 6 = - 2 + 6 = 4$

$T_{3} - T_{2} = 2 - - 2 = 2 + 2 = 4$

$i . e ., T_{2} - T_{1} = T_{3} - T_{2} = 4 = d$ , common difference

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

16th term, $T_{16} = 4 \times 16 - 10 = 64 - 10 = 54$

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