Question

the 6th term of an AP is 30 if its 20th term exceeds its 15th term by 25 ,find the Ap... guys solve this​

Answer 1

$\large\underline{Given:-}$

• Sum of 6th term of Ap ⇢ 30
• Sum of 20th term exceed 15th term by 25.

$\large\underline{To find:-}$

• find the Ap

$\large\underline{Solutions:-}$

$\: \: \: \: \: \fbox{{a_n} \: \: = \: \: {a} \: + \: {(n \: - \: 1)} \: d}$

• $\: \: \: \: \: {a} \: + \: {5d} \: \: = \: \: {30} \: \: \: \: \: .....{(1)}.$

Now,

✰ 20th term exceed 15th term by 25.

$\: \: \: \: \: \leadsto {a_{20}} \: - \: {a_{15}} \: \: = \: \: {25}$

$\: \: \: \: \: \leadsto {a} \: + \: {19d} \: - \: {({a} \: + \: {14d})} \: \: = \: \: {25}$

$\: \: \: \: \: \leadsto {a} \: + \: {19d} \: - \: {a} \: - \: {14d} \: \: = \: \: {25}$

$\: \: \: \: \: \leadsto {19d} \: - \: {14d} \: \: = \: \: {10}$

$\: \: \: \: \: \leadsto {5d} \: \: = \: \: {25}$

$\: \: \: \: \: \leadsto {d} \: \: = \: \: \frac{25}{5}$

$\: \: \: \: \: \leadsto {d} \: \: = \: \: {5}$

»★Then,

Putting value of d in Eq (1).

$\: \: \: \: \: \leadsto {a} \: + \: {5d} \: \: = \: \: {30}$

$\: \: \: \: \: \leadsto {a} \: + \: {5} \: \times \: {5} \: \: = \: \: {30}$

$\: \: \: \: \: \leadsto {a} \: + \: {25} \: \: = \: \: {30}$

$\: \: \: \: \: \leadsto {a} \: \: = \: \: {30} \: - \: {25}$

$\: \: \: \: \: \leadsto {a} \: \: = \: \: {5}$

✰ Ap ⇢ 5, 10, 15, 20....

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Answer 2

$\underline\bold{QUESTION}$

The 6th term of an AP is 39 if its 20th term exceeds its 15th term by 25 . Find the AP.

$\underline\bold{SOLUTION}$

$a+5d=30------(i)$

$a15+25= a20$

$=>a+14d+25=a+19d$

$=>25=19d-14d$

$=>25=5d$

$=>d=25/5$

$=>d=5$

Putting the value of d in equation(i)

$a+5d=30$

$a+5×5=30$

$a=30-25$

$a=5$

$AP= a,a+d,a+2d,a+3d.......$

$5,5+5,5+2×5,5+3×5.......$

$5,10,5+10,5+15......$

$5,10,15,20......$

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