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How many terms are there in 20, 25, 30......... 140

122
225
323
424

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Number of terms, \\ = {\frac{(1^{s t} t erm - last term)}{common difference}} + 1 \\ = (140 - \frac{20}{5}) + 1 \\ = (\frac{120}{5}) + 1 \\ = 24 + 1 \\ = 25 \end{aligned}

$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{terms}}, \cr & = \left\{ {\frac{{\left( {{1^{st}}\,t{\text{erm - last}}\,{\text{term}}} \right)}}{{{\text{common}}\,{\text{difference}}}}} \right\} + 1 \cr & = \left( {140 - \frac{{20}}{5}} \right) + 1 \cr & = \left( {\frac{{120}}{5}} \right) + 1 \cr & = 24 + 1 \cr & = 25 \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Number of terms, \\ = {\frac{(1^{s t} t erm - last term)}{common difference}} + 1 \\ = (140 - \frac{20}{5}) + 1 \\ = (\frac{120}{5}) + 1 \\ = 24 + 1 \\ = 25 \end{aligned}

$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{terms}}, \cr & = \left\{ {\frac{{\left( {{1^{st}}\,t{\text{erm - last}}\,{\text{term}}} \right)}}{{{\text{common}}\,{\text{difference}}}}} \right\} + 1 \cr & = \left( {140 - \frac{{20}}{5}} \right) + 1 \cr & = \left( {\frac{{120}}{5}} \right) + 1 \cr & = 24 + 1 \cr & = 25 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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