The 11th term of an A.P exceeds its 14th term by 14. Find the common difference.
Answers
AnswEr :
- Common difference is 2 .
GivEn :
- 11th term of an A.P exceeds its 14th term by 14 .
To find :
- Common difference (d) = ?
SoluTion :
According to the given information,
Therefore, common difference is 2.
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★ Formulae related to AP :
1] General form of an AP is,
- a, a + d, a + 2d, a + 3d and so on.
2] nth Term of an AP
- an = a + (n − 1) × d
3] Sum of N Terms of AP
- S = n/2[2a + (n − 1) × d]
Where,
- a = First term
- d = Common difference
- n = number of terms
- an = nth term
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Step-by-step explanation:
AnswEr :
Common difference is 2 .
\begin{gathered}\\ \\\end{gathered}
GivEn :
11th term of an A.P exceeds its 14th term by 14 .
\begin{gathered}\\ \\\end{gathered}
To find :
Common difference (d) = ?
\begin{gathered}\\ \\\end{gathered}
SoluTion :
\begin{gathered}\\\end{gathered}
According to the given information,
\begin{gathered}\\ \\\end{gathered}
\begin{gathered}: \implies \sf \: \: \: 11th \: term - 4th \: term =14 \\ \\ \\ \\\end{gathered}
:⟹11thterm−4thterm=14
\begin{gathered}: \implies \sf \: \: \: a+10d-(a+3d)=14 \\ \\ \\ \\\end{gathered}
:⟹a+10d−(a+3d)=14
\begin{gathered}: \implies \sf \: \: \: a+10d-a-3d=14 \\ \\ \\ \\\end{gathered}
:⟹a+10d−a−3d=14
\begin{gathered}: \implies \sf \: \: \: 7d=14 \\ \\ \\ \\\end{gathered}
:⟹7d=14
\begin{gathered}:\implies \sf \: \: \: { \underline{ \boxed{ \sf{ \purple{d=2}}}}} \: \bigstar \\ \\\end{gathered}
:⟹
d=2
★
Therefore, common difference is 2.
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\begin{gathered}\\\end{gathered}
★ Formulae related to AP :
\begin{gathered}\\\end{gathered}
1] General form of an AP is,
a, a + d, a + 2d, a + 3d and so on.
\begin{gathered}\\\end{gathered}
2] nth Term of an AP
an = a + (n − 1) × d
\begin{gathered}\\\end{gathered}
3] Sum of N Terms of AP
S = n/2[2a + (n − 1) × d]
\begin{gathered}\\\end{gathered}
Where,
a = First term
d = Common difference
n = number of terms
an = nth term
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