Question

if the first term of ap is 5 and its 100 term is 203 finds its 51th term

Answer 1

To Find :

• we need to find the 51th term of AP.

Solution :

• First term of AP = 5
• 100th term = 203

As we know that,

• an = a + (n - 1)d

›› a100 = a + (100 - 1)d

›› a100 = a + 99d

• 100th term = 203

›› a + 99d = 203

• First term (a) = 5

›› 5 + 99d = 203

›› 99d = 203 - 5

›› 99d = 198

›› d = 198/99

• ›› d = 2

So,

• Common difference (d) = 2

Now finding 51th term of AP :

››➔ a51 = a + (51 - 1)d

››➔ a51 = a + 50d

››➔ a51 = 5 + 50 × 2

››➔ a51 = 5 + 100

››➔ a51 = 105

Hence,

• 51th term of AP is 105 .

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Answer 2

$\huge\underline\blue{F{ \bf \pink{inD : }}}$

$\huge{51th\: { \red{T}erm}}$

$\huge\underline\green{G{\sf \red{iV\purple{eN}}}}$

$\huge{a = 5}$

$\huge{ a_{100} = 203}$

$\huge \sf \underline\red{{ \green{S}oLu \pink{Tion}}}$

$a_{100} = 203$

$a + (n - 1)d = 203$

$5 + (100 - 1)d = 203$

$5 + 99d = 203$

$99d = 203 - 5$

$99d = 198$

$d = \frac{198}{99} = 2$

$\huge\purple{formula \: to \: find \: a_{51}}$

$\huge \fbox{a + (n - 1)d}$

$where \: n = no. \: of \: term$

$d = common \: distance$

$a_{51} = 5 + (51 - 1)2$

$a_{51} = 5 + 50 \times 2$

$a_{51} = 5 + 100 = 105$

$\huge\bf\underline\green{ a_{51} = 105}$