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Question
In an A.P. the 10th term is 46 sum of the 5th and 7th term is 52. Find the A.P.
Solution
It is given that,
t10 = 46
t5 + t7 = 52
Now,
\[t_n = a + \left( n - 1 \right)d\]
\[ t_{10} = a + \left( 10 - 1 \right)d\]
\[ \Rightarrow 46 = a + 9d\]
\[ \Rightarrow a = 46 - 9d . . . \left( 1 \right)\]
\[ t_5 + t_7 = 52\]
\[ \Rightarrow \left( a + \left( 5 - 1 \right)d \right) + \left( a + \left( 7 - 1 \right)d \right) = 52\]
\[ \Rightarrow a + 4d + a + 6d = 52\]
\[ \Rightarrow 2a + 10d = 52\]
\[ \Rightarrow 2\left( 46 - 9d \right) + 10d = 52 \left( \text { from }\left( 1 \right) \right)\]
\[ \Rightarrow 92 - 18d + 10d = 52\]
\[ \Rightarrow 92 - 8d = 52\]
\[ \Rightarrow 8d = 92 - 52\]
\[ \Rightarrow 8d = 40\]
\[ \Rightarrow d = 5\]
\[ \Rightarrow a = 46 - 9\left( 5 \right) \left( \text { from }\left( 1 \right) \right)\]
\[ \Rightarrow a = 46 - 45\]
\[ \Rightarrow a = 1\]
Hence, the given A.P. is 1, 6, 11, 16, ....
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