For an AP show that Tp+2q …
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Sia ? 6 years, 4 months ago
Let the first term be a and the common difference be d.
we know, {tex}a_n= a+(n-1)d{/tex}
{tex}a_p=a+(p-1)d{/tex}
{tex}a_{p+2q}=a+(p+2q-1)d{/tex}
{tex}\therefore a_p+a_{p+2q}=a + (p - 1)d + a + (p + 2 q -1)d{/tex}
{tex}= a + pd - d + a + pd + 2qd - d{/tex}
{tex}= 2a + 2pd + 2qd - 2d{/tex}
{tex}= 2[a + (p + q - 1) d]{/tex} ............(i)
{tex}2a_{p+q}=2[a + (p + q - 1 ) d ]{/tex} ..........(ii)
From (i) and (ii), we get
ap + ap + 2q = 2ap+q
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