Question

12= The 8th term of can AP ES 17 and its 14th term
is -29. The common difference of the AP is
A) -2
B) 3
C) 2
D) 5​

Answer 1

$\rm\huge\red{\underline{\underline{ Question : }}}$

The 8th term of can AP is 17 and its 14th term is -29. The common difference of the AP is...?

$\rm\huge\red{\underline{\underline{ Solution : }}}$

★ Given that,

• Term 8 (a8) = 17
• Term 14 (a14) = - 29

★ To find,

• Common difference (d).

★ Let,

$\sf\:\implies a_{8} : a + 7d = 17 ..... (1)$

$\sf\:\implies a_{14} : a + 13d = - 29 ..... (1)$

• Subtract equations (1) & (2).We get,

$\sf\:\implies - 6d = 46$

$\sf\:\implies d = \frac{46}{ - 6}$

$\sf\:\implies d = -\frac{23}{3}$

$\underline{\boxed{\bf{\purple{\therefore Common\:difference\:(d) = -\frac{23}{3}}}}}\:\orange{\bigstar}$

★ More Information :

How to find common difference (d)?

Let us take an AP series.

AP : 1,2,3....

Let,

$\tt\:\implies a_{1} = 1$

$\tt\:\implies a_{2} = 2$

$\tt\:\implies a_{3} = 3$

Common difference (d) : a2 - a1 = a3 - a2

$\sf\:\implies 2 - 1 = 3 - 2$

$\sf\:\implies 1 = 1$

↪ From the above we can see that the difference between the successive terms is same (constant) which is 1.

↪ so we can say that the given sequence is in A.P.

↪ If the 1st term and the common difference 'd' is given then we can make an arithmetic sequence.

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$\boxed{\begin{minipage}{5 cm} AP Formulae \\ \\ : \implies a_{n} = a + (n - 1)d \\ \\ :\implies S_{n} = \frac{n}{2} [ 2a + (n - 1)d ] \end{minipage}}$

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Answer 2

To find common difference of an Ap.

EXPLANATION.

8th term of an Ap = 17

14th term of an Ap = -29

According to the question,

Nth term of an Ap

=> An = a + ( n - 1 ) d

8th term of an Ap = 17

=> a + 7d = 17 ..... (1)

14th term of an Ap = -29

=> a + 13d = -29 ..... (2)

From equation (1) and (2) we get,

=> -6d = 46

=> d = - 23 / 3

put the value of d = -23/3 in equation (1)

we get,

=> a + 7 X (-23) / 3 = 17

=> a - 161 / 3 = 17

=> a = 17 + 161 / 3

=> a = 51 + 161 / 3

=> a = 212 / 3

Therefore,

value of common difference =

-23/3

value of first term = 212/3