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The sum of $(p+q)^{\text {th }}$ and $(p-q)^{\text {th }}$ terms of an AP is equal to
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Twice the $\mathrm{p}^{\text {th }}$ term
Let the first term of $\mathrm{AP}=\mathrm{a}$ Let the common difference of $\mathrm{AP}=\mathrm{d}$ $(p+q)^{\text {th }}$ term $=\mathrm{T}_{\mathrm{ptq}}=\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d}$
$(p-q)^{\text {th }}$ term $=T_{p-q}^{p \cdot q}=a+(p-q-1) d$
$\mathrm{T}_{\mathrm{ptq}}+\mathrm{T}_{\mathrm{p-q}}=\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d}+\mathrm{a}+(\mathrm{p}-\mathrm{q}-1) \mathrm{d}$
$=2 a+(p+q-1+p-q-1) d$
$=2 a+(2 p-2) d=2[a+(p-1) d]=2 T_{p}$
$(p-q)^{\text {th }}$ term $=T_{p-q}^{p \cdot q}=a+(p-q-1) d$
$\mathrm{T}_{\mathrm{ptq}}+\mathrm{T}_{\mathrm{p-q}}=\mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d}+\mathrm{a}+(\mathrm{p}-\mathrm{q}-1) \mathrm{d}$
$=2 a+(p+q-1+p-q-1) d$
$=2 a+(2 p-2) d=2[a+(p-1) d]=2 T_{p}$
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