Question

# The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.                                                          [CBSE 2013]

Solution

## Let a be the first term and d be the common difference of the AP. Then, Also, From (1) and (2), we get $\frac{3d}{2}+8d=19\phantom{\rule{0ex}{0ex}}⇒\frac{3d+16d}{2}=19\phantom{\rule{0ex}{0ex}}⇒19d=38\phantom{\rule{0ex}{0ex}}⇒d=2$ Putting d = 2 in (1), we get $2a=3×2=6\phantom{\rule{0ex}{0ex}}⇒a=3$ So, ${a}_{2}=a+d=3+2=5\phantom{\rule{0ex}{0ex}}{a}_{3}=a+2d=3+2×2=7,...$ Hence, the AP is 3, 5, 7, 9, ... .MathematicsRS Aggarwal (2018)Standard X

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