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| 1. |
Determine the AP whose third term is 16 and 7th term exceeds the 5th term by 12. |
| Answer» Let the first term and the common difference of the AP be a and d respectively.Then,Third term = 16{tex} \\Rightarrow {/tex}\xa0a + (3 - 1)d = 16\xa0{tex}\\because {a_n} = a + (n - 1)d{/tex}{tex} \\Rightarrow {/tex}\xa0a + 2d = 16 ...... (1)and, 7th term = 5th term + 12{tex} \\Rightarrow {/tex}\xa0a + (7 - 1)d = a + (5 - 1)d + 12\xa0{tex}\\because {a_n} = a + (n - 1)d{/tex}{tex} \\Rightarrow {/tex}\xa0a + 6d = a + 4d + 12{tex} \\Rightarrow {/tex}\xa06d - 4d = 12{tex} \\Rightarrow {/tex}\xa02d = 12{tex} \\Rightarrow d = \\frac{{12}}{2} = 6{/tex}Put d = 6 in (1), we geta + 2(6) = 16{tex} \\Rightarrow {/tex}\xa0a + 12 = 16{tex} \\Rightarrow {/tex}\xa0a = 16 - 12{tex} \\Rightarrow {/tex}\xa0a = 4Hence, the required APs also4, 4 + 6, 4 + 6 + 6, 4 + 6 + 6 + 6, .......i.e. 4, 10, 16, 22, ....... | |