Question

Find the 31st term of AP whose 11th term is 38 and the 16th term is 73​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

we know that,

$\tt \red{{{a}}_{{n}} = a + (n - 1)d}$

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$\tt{Given \: {11}^{th} \: term \: is \: 38}$

$\tt{{{a}}_{{11}} = a + (11 - 1)d}$

$\tt{38 = a + (11 - 1)d}$

$\tt{38 = a + 10d}$

$\tt{38 - 10d = a}$

$\tt \blue{a = 38 - 10d} \: \: \: ...(1)$

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$\tt{Given \: {16}^{th} \: term \: is \: 73}$

$\tt{{{a}}_{{16}} = a + (16 - 1)d}$

$\tt{{{a}}_{{16}} = a +15d}$

$\tt{73 = a + 15d}$

$\tt{73 - 15d = a}$

$\tt \blue{a = 73 - 15d} \: \: \: ....(2)$

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Now,

From equation (1) & (2)

$\tt \blue{38 - 10d = 73 - 15d}$

$⟹ \tt \blue{38 - 73 = - 15d - 10d}$

$⟹ \tt \blue{ - 35 = - 5d}$

$⟹ \tt\blue{ \cancel\frac{ \cancel{ -} 35}{ \cancel{-} 5} = d}$

$⟹ \tt \blue{7 = d}$

$⟹ \tt \red{d = 7}$

Putting the value of d in equation (1)

$\tt{a = 38 = 10d}$

$\tt{a = 38 - 10 \times 7}$

$\tt{a = 38 - 70}$

$\tt \blue{a = - 32}$

We need to find the 31 st term

So, n = 31 , a = -32 , d = 7

$\tt{We \: need \: to \: find \: {{a}}_{{n}}}$

$\tt{Putting \: values}$

$\tt{{{a}}_{{31}} = - 32(31 - 1)7 }$

$\tt {= - 32 + 30 + 7}$

$\tt{ = - 32 + 210}$

$\tt{ = 178}$

$\tt \red{Therefore,the \: {31}^{st} \: term \: of \: AP \: is \: 178} \: Ans.$