Question

The sum of six terms which form an AP is 345.The difference between the first and the last term is 55.Find the terms.

Answer 1

Let the 6 terms be,

• $\displaystyle\sf {a_1=a-5d}$

• $\displaystyle\sf {a_2=a-3d}$

• $\displaystyle\sf {a_3=a-d}$

• $\displaystyle\sf {a_4=a+d}$

• $\displaystyle\sf {a_5=a+3d}$

• $\displaystyle\sf {a_6=a+5d}$

Since the 6 terms form an AP, the first term here is $\displaystyle\sf {a_1=a-5d}$ and the common difference is $\displaystyle\sf {2d.}$

The sum of these six terms is 345, i.e.,

$\displaystyle\longrightarrow\sf{\sum_{i=1}^6a_i=345}$

$\displaystyle\longrightarrow\sf{6a=345}$

$\displaystyle\longrightarrow\sf{a=57.5}$

Given that the difference between the first and the last terms is 55, i.e.,

$\displaystyle\longrightarrow\sf{a_6-a_1=55}$

$\displaystyle\longrightarrow\sf{(a+5d)-(a-5d)=55}$

$\displaystyle\longrightarrow\sf{a+5d-a+5d=55}$

$\displaystyle\longrightarrow\sf{10d=55}$

$\displaystyle\longrightarrow\sf{d=5.5}$

Hence the terms are,

$\displaystyle\longrightarrow\underline{\underline {\sf{a_1=57.5-5\times 5.5}=\bf{30}}}$

$\displaystyle\longrightarrow\underline{\underline {\sf{a_2=57.5-3\times 5.5}=\bf{41}}}$

$\displaystyle\longrightarrow\underline{\underline {\sf{a_3=57.5-5.5}=\bf{52}}}$

$\displaystyle\longrightarrow\underline{\underline {\sf{a_4=57.5+5.5}=\bf{63}}}$

$\displaystyle\longrightarrow\underline{\underline {\sf{a_5=57.5+3\times 5.5}=\bf{74}}}$

$\displaystyle\longrightarrow\underline{\underline {\sf{a_6=57.5+5\times 5.5}=\bf{85}}}$