Question

12. If 5th and 6th terms of an A.P. are respectively<br />6 and 5, find the 11th term of the A.P.​

Answer 1

★GIVEN★

If 5th and 6th terms of an A.P. are respectively 6 and 5.

★To Find★

The 11th term of the A.P.

★SOLUTION★

We know that,

$\large{\underline{\boxed{\sf{a_{n}=a+(n-1)d}}}}$

where,

• a is the first term of AP
• d is the common difference
• $\small{\bf{a_{n}\:is\:the\:nth\:term.}}$

It is said that 5th and 6th terms of an A.P. are respectively 6 and 5.

So,

$\large\implies{\sf{a_{5}=a+(n-1)d}}$

$\large\implies{\sf{6=a+(5-1)d}}$

$\large\implies{\sf{6=a+4d}}$

$\large\therefore\boxed{\sf{a+4d=6\red{----eq.(i)}}}$

$\large\implies{\sf{a_{6}=a+(n-1)d}}$

$\large\implies{\sf{5=a+(6-1)d}}$

$\large\implies{\sf{5=a+5d}}$

$\large\therefore\boxed{\sf{a+5d=5\red{----eq.(ii)}}}$

Now by subtracting eq.(ii) from eq.(i),

$\large{\sf{a+4d=6}}$---eq.(i)

$\large{\sf{a+5d=5}}$---eq.(ii)

$\large{\sf{(+)\cancel{a}\:\:\:\:\:\:+\:\:4d=(+)6}}$

$\large{\sf{(-)\cancel{a}+(-)5d=(-)5}}$_____________________________

$\large\implies{\sf{4d-5d=6-5}}$

$\large\implies{\sf{-d=1}}$

$\large\therefore\boxed{\sf{Common\:Difference(d)=-1}}$

Now putting d = -1 in eq.(i)

$\large\implies{\sf{a+4\times(-1)=6}}$

$\large\implies{\sf{a+(-4)=6}}$

$\large\implies{\sf{a-4=6}}$

$\large\implies{\sf{a=6+4}}$

$\large\therefore\boxed{\sf{a=10}}$

Now,

11th term of AP,

$\large\implies{\sf{a_{n}=a+(n-1)d}}$

$\large\implies{\sf{a_{11}=10+(11-1)(-1)}}$

$\large\implies{\sf{a_{11}=10+10\times(-1)}}$

$\large\implies{\sf{a_{11}=10+(-10)}}$

$\large\implies{\sf{a_{11}=10-10}}$

$\large\therefore\boxed{\sf{a_{11}=0}}$

$\large{\green{\underline{\boxed{\therefore{\bf{11th\:term\:of\:AP\:is\:0.}}}}}}$