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प्रश्न
The 7th term of the an AP is -4 and its 13th term is -16. Find the AP.
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उत्तर
We have
T7 = a + (n-1) d
⇒ a +6d = -4 ...............(1)
T13 = a +(n-1) d
⇒ a + 12d = -16 .................(2)
On solving (1) and (2), we get
a = 8 and d = - 2
Thus, first term = 8 and common difference = - 2
∴ The term of the AP are 8,6,4,2,.........
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Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
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