Question

If 11 AM's are inserted between $10$ and $28$, then -

• A. Number of integral AM's is $5$
• B. Difference between greatest and smallest AM is $15$
• C. Sum of greatest and smallest AM is $35$
• D. Sum of all AM's is $209$

Answer 1

The correct option is A Number of integral AM's is $5$

Initial term $\left(a\right)=10$

Final term $\left(l\right)=28$

11 AM's are inserted between 10 & 28

$\therefore$ therefore are 13 terms in sequence

last term (${13}^{th}$ term) = $10+(13-1)d$

$10+12d=28$

$12d=18;d=1.5$

Therefore the series will be

$a,a+d,a+2d,a+3d,a+4d,a+5d,a+6d,a+7d,a+8d,a+9d,a+10,a+11d,a+12d$

$10,11.5,13,14.5,16,17.5,19,20.5,22,23.5,25,26.5,28$

Therefore there are $5$ integral AM's