Question

Find p and q,such that 2p,2p+q,p+4q,35 are in A.P.

Answer 1

t2-t1=t3-t2=t4-t3
t1=2p
t2=2p+q
t3=p+4q
t4=35
2p+q-2p=p+4q-2p-q
q= 3q-p -equation 1
p+4q-2p-q =35-p -4q
7q=35
q=5
equation1
p=10

Answer 2

Answer:

$Value \: of \: p = 10\: and \: q = 5$

Step-by-step explanation:

$Given \: 2p,\:2p+q,\:p+4q,\:35 \: are \: in \:A.P$

/* we know that , "The difference between any two successive terms is the same throughout the series in A.P */

$i) t_{2}-t_{1} = t_{3}-t_{1}$

$\implies (2p+q) - 2p = (p+4q) - (2p+q)$

$\implies 2p+q-2p = p+4q-2p-q$

$\implies q = -p+3q$

$\implies q-3q = -p$

$\implies -2q = -p$

$\implies p = 2q\:---(1)$

$ii) t_{3}-t_{2}=t_{4}-t_{3}$

$\implies (p+4q)-(2p+q) = 35 - (p+4q)$

$\implies p+4q-2p-q = 35 - p - 4q$

$\implies -p+3q = 35 -p-4q$

$\implies 3q+4q = 35 - p + p$

$\implies 7q = 35$

$\implies q = \frac{35}{7}=5\:--(2)$

/* substitute q = 5 in equation (1) ,we get */

$p = 2\times 5 = 10$

Therefore.,

$Value \: of \: p = 10\: and \: q = 5$

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