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(p+q)th and (p-q)th term of an AP are respectively m and n. the p th term is​

Answer 1

Step-by-step explanation:

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student-name Elayanila asked in Math

(p+q) th term and( p-q)th term of an ap are respectively m and n. the pth term is

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student-name Sahil Sharma answered this

21 helpful votes in Math, Class XII-Science

a+(p+q-1)d=m

a+(p-q-1)d=n

subtracting both the equations we get

a-a+(p+q-1-p+q+1)d=m-n

0+(2q)d=m-n

d=(m-n)/2q

puting this value in the first equation we get

a+(p+q-1)(m-n)/2q=m

a+(p-1)(m-n)/2q + q(m-n)/2q =m

therefore a+(p-1)d +(m-n)/2 =m

so a+(p-1)d =m-(m-n)/2

=(2m-m+n)2

a+(p-1)d =(m+n)/2

Since a+(p-1)d is the form of the pth term therefore,

pth term = (m+n)/2

Answer 2

pth term is  $\dfrac{2mp-2m-mq+nq}{2p-2}}$

Step-by-step explanation:

Since we have given that

(p+q) th term of an AP = m

(p-q) th term of an AP = n

So, it becomes,

$a_{p+q}=a+(p+q-1)d=m\\\\a_{p-q}=a+(p-q-1)d=n$

so, it becomes,

$a+(p+q-1)d-[a-(p-q-1)d]=m-n\\\\a+(p+q-1)d-a+(p-q-1)d=m-n\\\\pd+qd-d+pd-qd-d=m-n\\\\2pd-2d=m-n\\\\2d(p-1)=m-n\\\\d=\dfrac{m-n}{2(p-1)}$

So,  first term would be

$a+(p+q-1)\times \dfrac{m-n}{2(p-1)}=m\\\\a=m-\dfrac{p+q-1(m-n)}{2p-2}\\\\a=\dfrac{2pm-2m-mp-mq+m+np+nq-n}{2p-2}\\\\a=\dfrac{mp-m-mq+np+nq-n}{2p-2}$

So, the value of pth term would be

$a+(p-1)d\\\\=\dfrac{mp-m-mq+np+nq-n}{2p-2}+(p-1)\dfrac{m-n}{2p-2}\\\\=\dfrac{mp-m-mq+np+nq-n+mp-np-m+n}{2p-2}\\\\=\dfrac{2mp-2m-mq+nq}{2p-2}}$

Hence, pth term is  $\dfrac{2mp-2m-mq+nq}{2p-2}}$

# learn more:

If tn denotes the nth term of the series 2+3+6+11+18+...........,then find t50.

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