Question

the 15th term of an ap exceeds its 8th term by 7 . find the common difference

Answer 1

Answer:

Let a be the first term and d the common difference.

It is given        

⇒ (a + 16d) - (a + 9d) = 7

⇒                          7d =7

⇒                            d = 1

Thus, the common difference is 7.

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\bf{Question}}}}}}}$

• the 15th term of an ap exceeds its 8th term by 7 . find the common difference

${\huge{\bold{\purple{\bigstar{\boxed{\bf{Answer}}}}}}}$

${\bold{\blue{\boxed{\bf{Given}}}}}$

• 15th term exceeds the 8th term by 7

${\bold{\blue{\boxed{\bf{To\:find}}}}}$

• $\bold {common\:difference= ???}$

${\bold{\blue{\boxed{\bf{ Formulae\:used }}}}}$

• $\bold a_n = a + (n-1)d$

${\bold{\blue{\boxed{\bf{ solution }}}}}$

• let the first term be a and common diff. be d

$\bold{ a_{15} = a + (15 -1)\times d }$

$\bold { a_{15} = a + 14 \times d }$

$\bold {a_{15} = a + 14d }$ ---- ( i )

$\bold {a_{8} = a + (8 - 1)\times d }$

$\bold { a_{8} = a + 7 \times d }$

$\bold {a_{8} = a + 7d }$ ---- ( ii )

• the 15th term of an ap exceeds its 8th term by 7 .

$\sf a_{15} - a_{8} = 7$

$\sf a + 14d - ( a + 7d ) = 7$

$\sf a + 14d - a - 7d = 7$

$\sf 7d = 7$

$\sf d = 1$

${\bold{\blue{\boxed{\boxed{\bf{common \: difference = 1 }}}}}}$

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