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1004 | QUES | The first 5 terms of a sequence are -18, -12, -6, 0, 6. If this an AP find 10th term: | Q | |
Given: | For a sequence, t1 = -18, t2 = -12, t3 = -6, t4 = 0, t5 = 6 | AS1 | ||
To Find: | Whether the sequence is an AP. If so, the 10th term i.e. t10. | AS2 | ||
Solution: | For the given sequence, | AS3 | ||
t2 - t1 = -12 - (-18) = -12 +18 | From given | AS4 | ||
∴ t2 - t1 = 6 | ….(1) | AS5 | ||
t3 - t2 = -6 - (-12) = -6 + 12 | From given | AS6 | ||
∴ t3 - t2 = 6 | ….(2) | AS7 | ||
t4 - t3 = 0 - (-6) = 0 + 6 | From given | AS8 | ||
∴ t4 - t3 = 6 | ….(3) | AS9 | ||
t5 - t4 = 6 - 0 = 6 | ….(4) | AS10 | ||
∴ t2 - t1 = t3 - t2 = t4 - t3 = t5 - t4 = 6 | from (1), (2), (3), (4) | AS11 | ||
∴ The difference between the consecutive terms is constant | AS12 | |||
∴ The given sequence is an AP with | AS13 | |||
the first term = a = -18 and common difference = d = 6 | ….(5) | AS14 | ||
Now For an AP, tn = a + (n - 1) d | AS15 | |||
∴ t10 = -18 + (10 - 1)(6) | Substituting values | AS16 | ||
∴ t10 = -18 + (9)(6) | AS17 | |||
∴ t10 = -18 + 54 | AS18 |
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