Let the three terms be a, a+d, a+2d. Given a(a+d)=120 and (a+d)(a+2d)=168. Testing a=10,d=2: 10×12=120 and 12×14=168. Both hold and d>0. Then a10=a+9d=10+18=28.
In an A.P., the product of the first term and the second term is 120 and the product of the second term and the third term is 168. Find the tenth term of the A.P. when common difference d > 0.
Held on 7 Jun 2023 · Verified 13 Jul 2026.
132
71
28
30
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