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| 1. |
in an AP if S5+S7=167 and S10=235 than find AP |
| Answer» \xa0According to the question, S5\xa0+\xa0S7\xa0= 167{tex}\\Rightarrow \\frac{5}{2}[2 a+4 d]+\\frac{7}{2}{/tex}[ 2a +\xa06d] = 167{tex}\\Rightarrow{/tex}\xa05(a + 2d) + 7(a + 3d) = 167{tex}\\Rightarrow{/tex}\xa012a + 31d = 167 ......(i)and S10\xa0= 235{tex}\\Rightarrow \\quad \\frac{10}{2}{/tex}[2a + 9d] = 235{tex}\\Rightarrow{/tex}\xa02a + 9d =\xa0{tex}\\frac{235}{5}{/tex}\xa0= 47 ........(ii)Multiplying eq. (ii) by 6 and then subtracting from (i), we haved =\xa0{tex}\\frac{-115}{-23}{/tex}\xa0= 5From (ii), we get2a + 9d = 47{tex}\\Rightarrow{/tex}\xa02a + 9(5) = 47{tex}\\Rightarrow{/tex}\xa02a = 47 - 45{tex}\\Rightarrow{/tex}\xa02a = 2\xa0{tex}\\Rightarrow{/tex}\xa0a = 1Therefore, A.P is 1, (1 + 5), (1 + 5 + 5), (1 + 5+ 5+ 5)........i.e, 1, 6, 11, 16...... | |