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प्रश्न
If the nth term of a progression is (4n – 10) show that it is an AP. Find its
(i) first term, (ii) common difference (iii) 16 the term.
उत्तर
`T_n = (4_n - 10 ) ` [Given]
`T_1 = (4xx 1 - 10 ) = -6`
`T_2 = (4xx2-10)=-2`
`T_3 = (4xx3 -10 ) =2`
`T_4= ( 4xx4-10) =6`
Clearly,[-2 -(-6)]=[2-(-2)] = [6-2] = 4 (Constant)
So, the terms - 6,-2,2,6,...... forms an AP.
Thus we have
(i) First term = - 6
(ii) Common difference = 4
(iii)` T_16 = a + ( n-1) d= a + 15d = -6 + 15 xx 4 =54`
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