Question

in an ap, the first term is 25 nth term is -17 and sum to first n terms is 60 find the common difference

Answer 1

$\sf Given \begin{cases} & \sf{First\;term,\;a = 25 } \\ & \sf{n^{th}\;term,\; a_n = -17} \\ & \sf{Sum\;of\;first\;n\;terms,\;S_n = 60 } </p><p>\end{cases}$

We have to find Common Difference, d?

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We know that,

$\star\;{\boxed{\sf{\purple{a_n = a + (n - 1)d}}}}$

$:\implies\sf - 17 = 25 + (n - 1)d$

$:\implies\sf (n - 1)d = - 42\qquad\qquad\bigg\lgroup\bf eq\;(1) \bigg\rgroup$

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Now,

$\star\;{\boxed{\sf{\purple{S_n = \dfrac{n}{2} \bigg( a + a_n \bigg)}}}}$

$\qquad\qquad:\implies\sf 60 = \dfrac{n}{2} \bigg( 25 + ( - 17) \bigg)$

$\qquad\qquad\quad:\implies\sf 60 \times 2 = 8n$

$\qquad\qquad\quad:\implies\sf n = \cancel{ \dfrac{120}{8}}$

$\qquad\qquad\quad:\implies{\boxed{\frak{\pink{n = 15}}}}\;\bigstar$

$\therefore$ There are 15 number of terms in given AP.

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⠀⠀⠀⠀ ⠀⠀✇ Now, Putting value of n in eq (1),

$\qquad\qquad\qquad:\implies\sf (15 - 1)d = - 42$

$\qquad\qquad\qquad\quad:\implies\sf 14d = - 42$

$\qquad\qquad\qquad\quad:\implies\sf d = - \cancel{ \dfrac{42}{14}}$

$\qquad\qquad\qquad\quad:\implies{\boxed{\frak{\pink{d = - 3}}}}\;\bigstar$

$\therefore$ Hence, Common Difference (d) between terms of AP is -3.

Answer 2

Answer:

sorry don't know nice dp hmm