Question

(11) The sixth term of an A.P. is 5 times the 1st term
and the eleventh term exceeds twice the fifth
term by 3. Find the 8th term.

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Answer 1

Given:

• The 6th term of an AP is 5 times the 1st term ⇒ $\sf a_6\ =\ 5a$

• The 11th term exceeds twice the 5th term by 3 ⇒ $\sf a_1_1\ =\ 2(a_5)\ +\ 3$

To find: The 8th term.

Answer:

$\sf a_6\ =\ 5a\\\\\\\implies a\ +\ 5d\ =\ 5a\\\\\\\implies\ 5d\ =\ 5a\ -\ a\\\\\\\implies\ 5d\ =\ 4a\\\\\\\implies\ d\ =\ \dfrac{4a}{5}$

Now,

$\sf a_1_1\ =\ 2(a_5)\ +\ 3\\\\\\\implies\ a\ +\ 10d\ =\ 2(a\ +\ 4d)\ +\ 3\\\\\\\implies\ a\ +\ 10d\ =\ 2a\ +\ 8d\ +\ 3\\\\\\\implies\ 10d\ -\ 8d\ =\ 2a\ -\ a\ +\ 3\\\\\\\implies\ 2d\ =\ a\ +\ 3$

Using the value of 'd' that we got earlier,

$\implies\ \sf 2\ \times\ \bigg(\dfrac{4a}{5}\bigg)\ =\ a\ +\ 3\\\\\\\implies\ \dfrac{8a}{5}\ =\ a\ +\ 3\\\\\\\implies\ 8a\ =\ 5(a\ +\ 3)\\\\\\\implies\ 8a\ =\ 5a\ +\ 15\\\\\\\implies\ 8a\ -\ 5a\ =\ 15\\\\\\\implies\ 3a\ =\ 15\\\\\\\implies\ a\ =\ \dfrac{15}{3}\\\\\\\implies\ \bf a\ =\ 5$

We now know the value of 'a'. Let's substitute that into the equational value of 'd' and find its actual value.

$\sf d\ =\ \dfrac{4a}{5}\\\\\\\implies\ d\ =\ \dfrac{4\ \times\ 5}{5}\\\\\\\implies\ d\ =\ \dfrac{20}{5}\\\\\\\implies\ \bf d\ =\ 4$

Since we know the values of 'a' and 'd', let's find the 8th term.

$\sf a_8\ =\ a\ +\ 7d\\\\\\a_8\ =\ 5\ +\ 7(4)\\\\\\a_8\ =\ 5\ +\ 28\\\\\\\bf a_8\ =\ 33$

Therefore, the 8th term is 33.